Thermodynamics – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral. Simulator   Previous Years AIEEE/JEE Mains Questions
Q. In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $\mathrm{CH}_{3} \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$ At 298 K standard Gibb’s energies of formation for $\mathrm{CH}_{3} \mathrm{OH}(\ell), \mathrm{H}_{2} \mathrm{O}(\ell)$ and $\mathrm{CO}_{2}(\mathrm{g})$ are –166.2, –237.2 and –394.4 kJ $\mathrm{mol}^{-1}$ respectively. If standard enthalpy of combustion of methanol is –726 kJ $\mathrm{mol}^{-1}$, efficiency of the fuel cell will be (1) 90% (2) 97% (3) 80% (4) 87% [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}+2 \mathrm{D}_{2} \mathrm{O}(l)$ $\Delta \mathrm{G}^{\circ}=\left(2 \Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{O}(l)+\Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left[\mathrm{CO}_{2}(g)\right]\right)-\left(\Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left[\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \Delta \mathrm{G}_{\mathrm{F}}^{\circ}\left[\mathrm{O}_{2}(g)\right]\right)\right.\right.$ $\Delta \mathrm{G}^{\circ}=(2(-237.2)+(-394.4)-(-166.2+0)$ $\Delta \mathrm{G}^{\mathrm{o}}=-868.8+166.2$ $=-702.6 \mathrm{kJ} / \mathrm{mol}$ Cell efficiency $=\left|\frac{\Delta \mathrm{G}^{\circ}}{\Delta \mathrm{H}^{\circ}}\right| \times 100$ $=\frac{702.6}{726} \times 100=96.77 \approx 97 \%$

Q. On the basis of the following thermochemical data : $\left(\Delta \mathrm{G}_{\mathrm{f}}^{0} \mathrm{H}_{(\mathrm{a}) \mathrm{y}}^{+}=0\right)$ $\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) ; \Delta \mathrm{H}=57.32 \mathrm{kJ}$ $\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286.20 \mathrm{kJ}$ The value of enthalpy of formation of $\mathrm{OH}^{-}$ ion at $25^{\circ} \mathrm{C}$ is :- (1) +228.88 kJ (2) –343.52 kJ (3) –22.88 kJ (4) –228.88 kJ [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}^{+}(\mathrm{aq} .)+\mathrm{OH}^{-}(\mathrm{aq} .)$ $57.32=\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\left[\mathrm{OH}^{-}(\mathrm{aq})\right]-\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\left[\mathrm{H}_{2} \mathrm{O}_{(l)}\right]$ $57.32=\Delta \mathrm{H}_{\mathrm{F}}^{\circ}\left[\mathrm{OH}_{(\mathrm{aq})}^{-}\right]+286.20$ $\Delta \mathrm{H}_{\mathrm{F}}^{\circ}\left[\mathrm{OH}^{-}_{(\mathrm{aq})}\right]=-228.88 \mathrm{kJ}$

Q. The standard enthalphy of formation of $\mathrm{NH}_{3}$ is $-46.0 \mathrm{kJ} \mathrm{mol}^{-1}$ If the enthalpy of formation of H2 from its atoms is –436 kJ $\mathrm{mol}^{-1}$ and that of $\mathrm{N}_{2}$ is $-712 \mathrm{kJ} \mathrm{mol}^{-1}$, the average bond enthalpy of N–H bond in NH3 is (1) $-1102 \mathrm{kJ} \mathrm{mol}^{-1}$ (2) $-964 \mathrm{kJ} \mathrm{mol}^{-1}$ (3) $+352$ kJ mol $^{-1}$ (4) $+1056 \mathrm{kJ} \mathrm{mol}^{-1}$ [AIEEE-2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})$ $-46=\frac{1}{2}(712)+\frac{3}{2}(436)-3(\mathrm{N}-\mathrm{H})$ – 46 = 356 + 654 – 3(N – H) N – H = 352 kJ / mol

Q. Consider the reaction : $4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}), \Delta_{\mathrm{r}} \mathrm{H}=-111 \mathrm{kJ}$ If $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{s})$ is formed instead of $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})$ in the above reaction, the rH value will be :- (given, of sublimation for $\mathrm{N}_{2} \mathrm{O}_{5}$ is 54 kJ \mathrm{mol}^{-1}) (1) –165 kJ (2) +54 kJ (3) +219 kJ (4) –219 kJ [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) If $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{s})$ is formed instead of $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})$ then $\Delta_{1} \mathrm{H}=-111-2(54)$ $\Delta_{\mathrm{r}} \mathrm{H}=-111-108$ $\Delta_{1} \mathrm{H}=-219 \mathrm{kJ} / \mathrm{mol}$

Q. The enthalpy of neutralisation of \mathrm{NH}_{4} \mathrm{OH}with HCl is –51.46 kJ \mathrm{mol}^{-1} and the enthalpy of neutralisation of NaOH with HCl is –55.90 kJ $\mathrm{mol}^{-1}$ The enthalpy of ionisation of $\mathrm{NH}_{4} \mathrm{OH}$ is: (1) $+107.36 \mathrm{kJ} \mathrm{mol}^{-1}$ (2) $-4.44 \mathrm{kJ} \mathrm{mol}^{-1}$ (3) $-107.36 \mathrm{kJ} \mathrm{mol}^{-1}$ (4) $+4.44 \mathrm{kJ} \mathrm{mo}$ [JEE-mains (online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The reaction $\mathrm{X} \rightarrow \mathrm{Y}$ is an exothermic reaction. Activation energy of the reaction for X into Y is 150 kJ $\operatorname{mol}^{-1}$. Enthalpy of reaction is 135 kJ $\mathrm{mol}^{-1}$ . The activation energy for the reverse reaction, $\mathrm{Y} \rightarrow \mathrm{X}$ will be : (1) 15 kJ $\mathrm{mol}^{-1}$ (2) 285 kJ $\mathrm{mol}^{-1}$ (3) 270 kJ $\mathrm{mol}^{-1}$ (4) 280 kJ $\mathrm{mol}^{-1}$

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\Delta \mathrm{H}=\mathrm{E}_{\mathrm{a}(\mathrm{f})}-\mathrm{E}_{\mathrm{a}(\mathrm{b})}$ $-135=150-\mathrm{E}_{\mathrm{a}(\mathrm{b})}$ – 617 = 161 + 520 + 77 + x – 1047 x = –328 kJ/mol

Q. Given Based on data provided, the value of electron gain enthalpy of fluorine would be : (1) –300 kJ $\mathrm{mol}^{-1}$ (2) –328 kJ $\mathrm{mol}^{-1}$ (3) –350 kJ $\mathrm{mol}^{-1}$ (4) –228 kJ $\mathrm{mol}^{-1}$ [JEE-mains (online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Given : The molar enthalpy of vapourisation of water will be :- (1) 241. 8 kJ $\mathrm{mol}^{-1}$ (2) 527.7 kJ $\mathrm{mol}^{-1}$ (3) 44.1 kJ $\mathrm{mol}^{-1}$ (4) 22.0 kJ $\mathrm{mol}^{-1}$ [JEE-mains (online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}_{298}^{\circ}\right)$ for methane, $\mathrm{CH}_{4}$ is– 74.9 kJ $\mathrm{mol}^{-1}$. In order to calculate the average energy given out in the formation of a C–H bond from this it is necessary to know which one of the following? (1) the dissociation energy of the hydrogen molecule, $\mathrm{H}_{2}$. (2) the dissociation energy of $\mathrm{H}_{2}$ and enthalpy of sublimation of carbon (graphite). (3) the first four ionisation energies of carbon and electron affinity of hydrogen. (4) the first four ionisation energies of carbon. [JEE-mains(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) From formation enthalpy of methane C-H bond enthalpy can be calculated as – $\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})$ $\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{CH}_{4}(\mathrm{g})\right)=\Delta \mathrm{H}_{\mathrm{a}}^{\circ}[\mathrm{C}(\mathrm{s})]+2(\mathrm{H}-\mathrm{H})-4(\mathrm{C}-\mathrm{H})$ $(\mathrm{C}-\mathrm{H})=\frac{\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{CH}_{4}(\mathrm{g})\right)-\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{C}(\mathrm{s}))-2[\mathrm{H}-\mathrm{H}]}{4}$ So for calculating (C-H) bond we will required data of H- H bond enthalpy and sublimation enthalpy of carbon solid.

Q. For an ideal Solution of two components A and B, which of the following is true ? (1) $\Delta \mathrm{H}_{\text {mixing }}<0$ (zero) (2) A – A, B – B and A – B interactions are identical (3) A – B interaction is stronger than A – A and B – B interactions (4) $\Delta \mathrm{H}_{\text {mixing }}>0$ (zero) [JEE-mains (online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. For complete combustion of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\ell)$the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ $\mathrm{mol}^{-1}$ at $25^{\circ} \mathrm{C}$. Assuming ideality the Enthalpy of combustion, $\Delta_{\mathrm{C}} \mathrm{H}$, for the raction will be :- $\left(\mathrm{R}=8.314 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ (1) $-1460.50 \mathrm{kj} \mathrm{mol}^{-1}$ (2) $-1350.50 \mathrm{kJ} \mathrm{mol}^{-1}$ (3) $-1366.95 \mathrm{kJ} \mathrm{mol}^{-1}$ $(4)-1361.95 \mathrm{kJ} \mathrm{mol}^{-1}$ [JEE-mains(offline)2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The heats of combustion of carbon and carbon monoxide are – 393.5 and – 285.5 kJ $\operatorname{mol}^{-1}$, respectively. The heat of formation (in kJ) of carbon monoxide per mole is : (1)– 110 (2) 110.5 (3) 676.5 (4) – 676.5 [JEE-Mains 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Given [JEE – Main – 2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) ; \Delta \mathrm{H}^{\circ}=890.3$ $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}-393.5-285.8$ $\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=\sum\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)_{\text {protucts }}-\sum\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)_{\text {Restatatis }}$ $890.3=\left[1 \times\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)_{\mathrm{CH}_{4}}+2 \times 0\right]-[1 \times(-393.5)+2(-285.8)]$ $\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)_{\mathrm{CH}_{4}}=890.3-965.1=-74.8 \mathrm{kJ} / \mathrm{mol}$

Q. The combustion of benzene (l) gives $\mathrm{CO}_{2}(\mathrm{g})$ and $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$. Given that heat of combustion of benzene at constant volume is –3263.9 kJ $\operatorname{mol}^{-1}$ at $25^{\circ} \mathrm{C}$; heat of combustion (in kJ $\mathrm{mol}^{-1}$) of benzene at constant pressure will be – (R = 8.314 $\mathrm{JK}^{-1}$ $\mathrm{mol}^{-1}$) (1)–452.46 (2) 3260 (3) –3267.6 (4) 4152.6 [JEE – Main – 2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Themochemistry – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral. Simulator   Previous Years AIEEE/JEE Mains Questions
Q. In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $\mathrm{CH}_{3} \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$ At 298 K standard Gibb’s energies of formation for $\mathrm{CH}_{3} \mathrm{OH}(\ell), \mathrm{H}_{2} \mathrm{O}(\ell)$ and $\mathrm{CO}_{2}(\mathrm{g})$ are –166.2, –237.2 and –394.4 kJ $\mathrm{mol}^{-1}$ respectively. If standard enthalpy of combustion of methanol is –726 kJ $\mathrm{mol}^{-1}$, efficiency of the fuel cell will be (1) 90% (2) 97% (3) 80% (4) 87% [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}+2 \mathrm{D}_{2} \mathrm{O}(l)$ $\Delta \mathrm{G}^{\circ}=\left(2 \Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{O}(l)+\Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left[\mathrm{CO}_{2}(g)\right]\right)-\left(\Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left[\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \Delta \mathrm{G}_{\mathrm{F}}^{\circ}\left[\mathrm{O}_{2}(g)\right]\right)\right.\right.$ $\Delta \mathrm{G}^{\circ}=(2(-237.2)+(-394.4)-(-166.2+0)$ $\Delta \mathrm{G}^{\mathrm{o}}=-868.8+166.2$ $=-702.6 \mathrm{kJ} / \mathrm{mol}$ Cell efficiency $=\left|\frac{\Delta \mathrm{G}^{\circ}}{\Delta \mathrm{H}^{\circ}}\right| \times 100$ $=\frac{702.6}{726} \times 100=96.77 \approx 97 \%$

Q. On the basis of the following thermochemical data : $\left(\Delta \mathrm{G}_{\mathrm{f}}^{0} \mathrm{H}_{(\mathrm{a}) \mathrm{y}}^{+}=0\right)$ $\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) ; \Delta \mathrm{H}=57.32 \mathrm{kJ}$ $\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286.20 \mathrm{kJ}$ The value of enthalpy of formation of $\mathrm{OH}^{-}$ ion at $25^{\circ} \mathrm{C}$ is :- (1) +228.88 kJ (2) –343.52 kJ (3) –22.88 kJ (4) –228.88 kJ [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}^{+}(\mathrm{aq} .)+\mathrm{OH}^{-}(\mathrm{aq} .)$ $57.32=\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\left[\mathrm{OH}^{-}(\mathrm{aq})\right]-\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\left[\mathrm{H}_{2} \mathrm{O}_{(l)}\right]$ $57.32=\Delta \mathrm{H}_{\mathrm{F}}^{\circ}\left[\mathrm{OH}_{(\mathrm{aq})}^{-}\right]+286.20$ $\Delta \mathrm{H}_{\mathrm{F}}^{\circ}\left[\mathrm{OH}^{-}_{(\mathrm{aq})}\right]=-228.88 \mathrm{kJ}$

Q. The standard enthalphy of formation of $\mathrm{NH}_{3}$ is $-46.0 \mathrm{kJ} \mathrm{mol}^{-1}$ If the enthalpy of formation of H2 from its atoms is –436 kJ $\mathrm{mol}^{-1}$ and that of $\mathrm{N}_{2}$ is $-712 \mathrm{kJ} \mathrm{mol}^{-1}$, the average bond enthalpy of N–H bond in NH3 is (1) $-1102 \mathrm{kJ} \mathrm{mol}^{-1}$ (2) $-964 \mathrm{kJ} \mathrm{mol}^{-1}$ (3) $+352$ kJ mol $^{-1}$ (4) $+1056 \mathrm{kJ} \mathrm{mol}^{-1}$ [AIEEE-2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})$ $-46=\frac{1}{2}(712)+\frac{3}{2}(436)-3(\mathrm{N}-\mathrm{H})$ – 46 = 356 + 654 – 3(N – H) N – H = 352 kJ / mol

Q. Consider the reaction : $4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}), \Delta_{\mathrm{r}} \mathrm{H}=-111 \mathrm{kJ}$ If $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{s})$ is formed instead of $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})$ in the above reaction, the rH value will be :- (given, of sublimation for $\mathrm{N}_{2} \mathrm{O}_{5}$ is 54 kJ \mathrm{mol}^{-1}) (1) –165 kJ (2) +54 kJ (3) +219 kJ (4) –219 kJ [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) If $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{s})$ is formed instead of $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})$ then $\Delta_{1} \mathrm{H}=-111-2(54)$ $\Delta_{\mathrm{r}} \mathrm{H}=-111-108$ $\Delta_{1} \mathrm{H}=-219 \mathrm{kJ} / \mathrm{mol}$

Q. The enthalpy of neutralisation of \mathrm{NH}_{4} \mathrm{OH}with HCl is –51.46 kJ \mathrm{mol}^{-1} and the enthalpy of neutralisation of NaOH with HCl is –55.90 kJ $\mathrm{mol}^{-1}$ The enthalpy of ionisation of $\mathrm{NH}_{4} \mathrm{OH}$ is: (1) $+107.36 \mathrm{kJ} \mathrm{mol}^{-1}$ (2) $-4.44 \mathrm{kJ} \mathrm{mol}^{-1}$ (3) $-107.36 \mathrm{kJ} \mathrm{mol}^{-1}$ (4) $+4.44 \mathrm{kJ} \mathrm{mo}$ [JEE-mains (online) 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The reaction $\mathrm{X} \rightarrow \mathrm{Y}$ is an exothermic reaction. Activation energy of the reaction for X into Y is 150 kJ $\operatorname{mol}^{-1}$. Enthalpy of reaction is 135 kJ $\mathrm{mol}^{-1}$ . The activation energy for the reverse reaction, $\mathrm{Y} \rightarrow \mathrm{X}$ will be : (1) 15 kJ $\mathrm{mol}^{-1}$ (2) 285 kJ $\mathrm{mol}^{-1}$ (3) 270 kJ $\mathrm{mol}^{-1}$ (4) 280 kJ $\mathrm{mol}^{-1}$

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\Delta \mathrm{H}=\mathrm{E}_{\mathrm{a}(\mathrm{f})}-\mathrm{E}_{\mathrm{a}(\mathrm{b})}$ $-135=150-\mathrm{E}_{\mathrm{a}(\mathrm{b})}$ – 617 = 161 + 520 + 77 + x – 1047 x = –328 kJ/mol

Q. Given Based on data provided, the value of electron gain enthalpy of fluorine would be : (1) –300 kJ $\mathrm{mol}^{-1}$ (2) –328 kJ $\mathrm{mol}^{-1}$ (3) –350 kJ $\mathrm{mol}^{-1}$ (4) –228 kJ $\mathrm{mol}^{-1}$ [JEE-mains (online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Given : The molar enthalpy of vapourisation of water will be :- (1) 241. 8 kJ $\mathrm{mol}^{-1}$ (2) 527.7 kJ $\mathrm{mol}^{-1}$ (3) 44.1 kJ $\mathrm{mol}^{-1}$ (4) 22.0 kJ $\mathrm{mol}^{-1}$ [JEE-mains (online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}_{298}^{\circ}\right)$ for methane, $\mathrm{CH}_{4}$ is– 74.9 kJ $\mathrm{mol}^{-1}$. In order to calculate the average energy given out in the formation of a C–H bond from this it is necessary to know which one of the following? (1) the dissociation energy of the hydrogen molecule, $\mathrm{H}_{2}$. (2) the dissociation energy of $\mathrm{H}_{2}$ and enthalpy of sublimation of carbon (graphite). (3) the first four ionisation energies of carbon and electron affinity of hydrogen. (4) the first four ionisation energies of carbon. [JEE-mains(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) From formation enthalpy of methane C-H bond enthalpy can be calculated as – $\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})$ $\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{CH}_{4}(\mathrm{g})\right)=\Delta \mathrm{H}_{\mathrm{a}}^{\circ}[\mathrm{C}(\mathrm{s})]+2(\mathrm{H}-\mathrm{H})-4(\mathrm{C}-\mathrm{H})$ $(\mathrm{C}-\mathrm{H})=\frac{\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{CH}_{4}(\mathrm{g})\right)-\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{C}(\mathrm{s}))-2[\mathrm{H}-\mathrm{H}]}{4}$ So for calculating (C-H) bond we will required data of H- H bond enthalpy and sublimation enthalpy of carbon solid.

Q. For an ideal Solution of two components A and B, which of the following is true ? (1) $\Delta \mathrm{H}_{\text {mixing }}<0$ (zero) (2) A – A, B – B and A – B interactions are identical (3) A – B interaction is stronger than A – A and B – B interactions (4) $\Delta \mathrm{H}_{\text {mixing }}>0$ (zero) [JEE-mains (online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. For complete combustion of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\ell)$the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ $\mathrm{mol}^{-1}$ at $25^{\circ} \mathrm{C}$. Assuming ideality the Enthalpy of combustion, $\Delta_{\mathrm{C}} \mathrm{H}$, for the raction will be :- $\left(\mathrm{R}=8.314 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ (1) $-1460.50 \mathrm{kj} \mathrm{mol}^{-1}$ (2) $-1350.50 \mathrm{kJ} \mathrm{mol}^{-1}$ (3) $-1366.95 \mathrm{kJ} \mathrm{mol}^{-1}$ $(4)-1361.95 \mathrm{kJ} \mathrm{mol}^{-1}$ [JEE-mains(offline)2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The heats of combustion of carbon and carbon monoxide are – 393.5 and – 285.5 kJ $\operatorname{mol}^{-1}$, respectively. The heat of formation (in kJ) of carbon monoxide per mole is : (1)– 110 (2) 110.5 (3) 676.5 (4) – 676.5 [JEE-Mains 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Given [JEE – Main – 2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) ; \Delta \mathrm{H}^{\circ}=890.3$ $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}-393.5-285.8$ $\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=\sum\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)_{\text {protucts }}-\sum\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)_{\text {Restatatis }}$ $890.3=\left[1 \times\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)_{\mathrm{CH}_{4}}+2 \times 0\right]-[1 \times(-393.5)+2(-285.8)]$ $\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)_{\mathrm{CH}_{4}}=890.3-965.1=-74.8 \mathrm{kJ} / \mathrm{mol}$

Q. The combustion of benzene (l) gives $\mathrm{CO}_{2}(\mathrm{g})$ and $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$. Given that heat of combustion of benzene at constant volume is –3263.9 kJ $\operatorname{mol}^{-1}$ at $25^{\circ} \mathrm{C}$; heat of combustion (in kJ $\mathrm{mol}^{-1}$) of benzene at constant pressure will be – (R = 8.314 $\mathrm{JK}^{-1}$ $\mathrm{mol}^{-1}$) (1)–452.46 (2) 3260 (3) –3267.6 (4) 4152.6 [JEE – Main – 2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

States of Matter – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral. Simulator   Previous Years AIEEE/JEE Mains Questions
Q. The molecular velocity of any gas is :- (1) inversely proportional to the square root of temperature (2) inversely proportional to absolute temperature (3) directly proportional to square of temperature (4) directly proportional to square root of temperature [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. , v and u represent most probable velocity, average velocity and root mean square velocity respectively of a gas at a particular temperature. The correct order among the following is – (1)  > u > v (2) v > u >  (3) u > v >  (4) u >  > v [JEE(Main)-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. An open vessel at 300 K is heated till $\frac{2}{5}$th of the air in it is expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated is :- (1) 750 K (2) 400 K (3) 500 K (4) 1500K [JEE(Main-online)-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. For 1 mol of an ideal gas at constant temperature T, the plot of (log P) against(logV) is a (P : Pressure, V : Volume) :- (1) Straight line parallel to x-axis (2) Curve starting at origin] (3) Straight line with a negative slope (4) Straight line passing through origin [JEE(Main-online)-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) PV = nRT log (PV) = log (nRT) log P + log V = log (nRT) log P = –log V + log (nRT) y = – mx + C slope $=\tan \theta=-1$

Q. The relationship among most probable velocity, average velocity and root mean square velocity is respectively :- (1) $\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}$ (2) $\sqrt{2}: \sqrt{3}: \sqrt{8 / \pi}$ (3) $\sqrt{3}: \sqrt{8 / \pi}: \sqrt{2}$ (4) $\sqrt{8 / \pi}: \sqrt{3}: \sqrt{2}$ [JEE(Main-online)-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Which one of the following is the wrong assumption of kinetic theory of gases ? (1) All the molecules move in straight line between collision and with same velocity. (2) Molecules are separated by great distances compared to their sizes. (3) Pressure is the result of elastic collision of molecules with the container’s wall. (4) Momentum and energy always remain conserved. [JEE(Main-online)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecules in a container of fixed volume is increased from $5 \times 10^{4} \mathrm{cm} /$ s to $10 \times 10^{4} \mathrm{cm} / \mathrm{s} ?$ (1) Four (2) three (3) Two (4) Six [JEE(Main-online)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{u}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}$ $\left(\frac{5 \times 10^{4}}{10 \times 10^{4}}\right)=\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}$ $\frac{1}{4}=\frac{T_{1}}{T_{2}}$ $\mathrm{T}_{2}=4 \mathrm{T}_{1}$

Q. For gaseous state, if most probable speed is denoted by C, average speed by $\overline{\mathrm{c}}$ and mean square speed by C, then for a large number of molecules the ratios of these speeds are :- (1) $\mathrm{C}: \overline{\mathrm{C}}: \mathrm{C}=1.225: 1.128: 1$ (2) $\mathrm{C}: \overline{\mathrm{C}}: \mathrm{C}=1.128: 1.225: 1$ (3) $\mathrm{C}: \overline{\mathrm{C}}: \mathrm{C}=1: 1.128: 1.225$ (4) $\mathrm{C}: \overline{\mathrm{C}}: \mathrm{C}=1: 1.225: 1.128$ [JEE(Main-offline)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\mathrm{u}_{\mathrm{avg}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ $\mathrm{u}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{u}_{\mathrm{mp}}=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$

Q. A gaseous compound of nitrogen and hydrogen contains 12.5%(by mass) of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is : (1) $\mathrm{NH}_{2}$ (2) $\mathrm{NH}_{3}$ (3) $\mathrm{N}_{3} \mathrm{H}$ (4) $\mathrm{N}_{2} \mathrm{H}_{4}$ [JEE(Main-online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The initial volume of a gas cylinder is 750.0 mL. If the pressure of gas inside the cylinder changes from 840.0 mm Hg to 360.0 mm Hg, the final volume the gas will be (1)1.750 L (2) 7.50 L (3) 3.60 L (4) 4.032 L [JEE(Main-online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}$ $840 \times 750=360 \times \mathrm{V}_{2}$ $\mathrm{V}_{2}=\frac{840 \times 750}{360}$

Q. The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is : (Atomic masses : He = 4 u, O = 16 u) (1) 1200 K (2) 600 K (3) 300 K (4) 2400 K [JEE(Main-online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Two closed bulbs of equal volume(V) containing an ideal gas initially at pressure pi and temperature $\mathrm{T}_{1}$ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to $\mathrm{T}_{2}$. The final pressure $\mathrm{P}_{\mathrm{f}}$ is :- [JEE(Main)-2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Initial moles and final moles are equal $\left(\mathrm{n}_{\mathrm{T}}\right)_{\mathrm{i}}=\left(\mathrm{n}_{\mathrm{T}}\right)_{\mathrm{f}}$ $\frac{\mathrm{P}_{\mathrm{i}} \mathrm{V}}{\mathrm{RT}_{1}}+\frac{\mathrm{P}_{\mathrm{i}} \mathrm{V}}{\mathrm{RT}_{1}}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{V}}{\mathrm{RT}_{1}}+\frac{\mathrm{P}_{\mathrm{f}} \mathrm{V}}{\mathrm{RT}_{2}}$ $2 \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{T}_{\mathrm{i}}}=\frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{T}_{\mathrm{i}}}+\frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{T}_{2}}$ $\mathrm{P}_{\mathrm{f}}=\frac{2 \mathrm{P}_{\mathrm{i}} \mathrm{T}_{2}}{\mathrm{T}_{1}+\mathrm{T}_{2}}$

Q. ‘a’ and ‘b’ are Vander Waal’s constants for gases. Chlorine is more easily liquefied than ethane because :- [Aieee-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Higher the ‘a’ value, more easily the gas is liquified, lower the ‘b’ value, more easily the gas is liquified

Q. The compressibility factor for a real gas at high pressure is :- (1) $1-\frac{\text { Pb }}{\mathrm{RT}}$ (2) $1+\frac{\mathrm{RT}}{\mathrm{Pb}}$ (3) 1 (4) $1+\frac{\mathrm{Pb}}{\mathrm{RT}}$ [Aieee-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) At high pressure, repulsion dominate. $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{Vm}^{2}}\right)(\mathrm{Vm}-\mathrm{b})=\mathrm{RT}$ P(Vm – b) = RT $\mathrm{So}, \mathrm{Z}=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$

Q. If Z is the compressibility factor, van der Waals’ equation at low pressure can be written as : $(1) Z=1-\frac{\mathrm{Pb}}{\mathrm{RT}}$ $(2) Z=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$ (3) $\mathrm{Z}=1+\frac{\mathrm{RT}}{\mathrm{Pb}}$ (4) $\mathrm{Z}=1-\frac{\mathrm{a}}{\mathrm{V}_{\mathrm{m}} \mathrm{RT}}$ [JEE-MAINS-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) At low pressure, attractions dominate. $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{Vm}^{2}}\right)(\mathrm{Vm}-\mathrm{b})=\mathrm{RT}$ $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{Vm}^{2}}\right)(\mathrm{Vm})=\mathrm{RT}$ $\mathrm{So}, \mathrm{Z}=1-\frac{\mathrm{a}}{\mathrm{VmRT}}$

Q. When does a gas deviate the most from it’s ideal behaviour ? (1) At high pressure and low temperature (2) At high pressure and high temperature (3) At low pressure and low temperature (4) At low pressure and high temperature [JEEMAINS(online)-2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) Gas behaves most ideally at high T and low P while deviates most from ideal behaviour at high P and low T.

Redox Reaction – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.   Simulator   Previous Years AIEEE/JEE Mains Questions
Q. Consider the following reaction: $\mathrm{xMnO}_{4}^{-}+\mathrm{yC}_{2} \mathrm{O}_{4}^{2-}+\mathrm{ZH}^{+} \rightarrow$ $\mathrm{xMn}^{2+}+2 \mathrm{yCO}_{2}+\frac{\mathrm{Z}}{2} \mathrm{H}_{2} \mathrm{O}$ The values of x, y and z in the reaction are respectively :- (1) 5,2 and 16 (2) 2,5 and 8 (3) 2, 5 and 16 (4) 5,2 and 8 [JEE(Main)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) Fact

Q. Given : $\mathrm{X} \mathrm{Na}_{2} \mathrm{HAsO}_{3}+\mathrm{Y} \mathrm{NaBrO}_{3}+\mathrm{ZHCl} \rightarrow \mathrm{NaBr}$ $+\mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{NaCl}$ The values of X, Y and Z in the above redox reaction are respectively : (1) 2, 1, 3 (2) 3, 1, 6 (3) 2, 1, 2 (4) 3, 1, 4 [JEE(Main-online)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Fact

Q. In which of the following reaction $\mathrm{H}_{2} \mathrm{O}_{2}$ acts as a reducing agent ? (1) $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}$ (2) $\mathrm{H}_{2} \mathrm{O}_{2}-2 \mathrm{e}^{-} \rightarrow \mathrm{O}_{2}+2 \mathrm{H}^{+}$ (3) $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{OH}^{-}$ (4) $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{OH}^{-}-2 \mathrm{e}^{-} \rightarrow \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O}$ (1) (1), (3) (2) (2), (4) (3) (1), (2) (4) (3), (4) [JEE(Main-online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Fact

Q. Consider the reaction $\mathrm{H}_{2} \mathrm{SO}_{3(\mathrm{aq})}+\mathrm{Sn}_{(\mathrm{aq})}^{4+}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{Sn}_{(\mathrm{aq})}^{2+}+\mathrm{HSO}_{4(\mathrm{aq})}^{-}+3 \mathrm{H}_{(\mathrm{aq})}^{+}$ Which of the following statements is correct? (1) $\mathrm{H}_{2} \mathrm{SO}_{3}$ is the reducing agent because it undergoes oxidation (2) $\mathrm{H}_{2} \mathrm{SO}_{3}$ is the reducing agent because it undergoes reduction (3) $\mathrm{Sn}^{4+}$ is the reducing agent because it undergoes oxidation (4) $\mathrm{Sn}^{4+}$ is the oxidizing agent because it undergoes oxidation [JEE(Main-online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) Fact

Q. How many electrons are involved in the following redox reaction ? $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+}+\mathrm{CO}_{2}$ (Unbalanced) (1) 3 (2) 4 (3) 5 (4) 6 [JEE(Main-online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Fact

Q. Which of the following reactions is an example of a redox reaction ? (1) $\mathrm{XeF}_{4}+\mathrm{O}_{2} \mathrm{F}_{2} \rightarrow \mathrm{XeF}_{6}+\mathrm{O}_{2}$ (2) $\mathrm{XeF}_{2}+\mathrm{PF}_{5} \rightarrow[\mathrm{XeF}]^{+} \mathrm{PF}_{6}^{-}$ (3) $\mathrm{XeF}_{6}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{XeOF}_{4}+2 \mathrm{HF}$ (4) $\mathrm{XeF}_{6}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{XeO}_{2} \mathrm{F}_{2}+4 \mathrm{HF}$ [JEE-Main 2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) In the reaction $\stackrel{+4}{\mathrm{X}} \mathrm{e} \mathrm{F}_{4}+\mathrm{O}_{2} \mathrm{F}_{2} \rightarrow \stackrel{+6}{\mathrm{X}} \mathrm{eF}_{6}+\mathrm{O}_{2}^{0}$ Xenon undergoes oxidation while oxygen undergoes reduction.

Q. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination ? [JEE-Main 2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Methyl orange shows Red(pinkish) color in Acidic medium & yellow color in basic medium since original solution is basic so initial color  yellow & Titrated with acid so Final color  pinkish (red)

Mole Concept – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral. Simulator Previous Years AIEEE/JEE Mains Questions
Q. A 5.2 molal aqueous solution of methyl alcohol, $\mathrm{CH}_{3} \mathrm{OH}$, is supplied. What is the mole fraction of methyl alcohol in the solution ? (1) 0.086           (2) 0.050             (3) 0.100             (4) 0.190 [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $5.2 \mathrm{m} \mathrm{CH}_{3} \mathrm{OH}$ ie $5.2 \mathrm{mol} \mathrm{CH}_{3} \mathrm{OH}$ present in $1 \mathrm{kg}$ water $\mathrm{X}_{\mathrm{M}}=\frac{5.2}{5.2+\frac{1000}{18}}=\frac{5.2}{5.2+55.5}=\frac{5.2}{60.7}=.086$

Q. The concentrated sulphuric acid that is peddled commercially is 95% $\mathrm{H}_{2} \mathrm{SO}_{4}$ by weight. If the density of this commerical acid is 1.834 g $\mathrm{cm}^{-3}$, the molarity of this solution is :- (1) 17.8 M         (2) 15.7 M           (3) 10.5 M           (4) 12.0 M [aieee-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $95 \% \frac{\mathrm{w}}{\mathrm{w}} \mathrm{H}_{2} \mathrm{SO}_{4}$ i.e. $100 \mathrm{gm}$ contain $95 \mathrm{gm} \mathrm{H}_{2} \mathrm{SO}_{4}$ Molarity $=\frac{95 / 98}{100 / 1.834} \times 1000$ $=\frac{95 \times 1.834}{98 \times 100} \times 1000=1.78$

Q. The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is (1) 2.05 M           (2) 0.50 M            (3) 1.78 M             (4) 1.02 M [AIEEE-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) Mass of Solution = 120 + 1000 = 1120 gm Vol of solution $=\frac{\text { mass }}{\text { density }}$ $=\frac{1120}{1.15} \mathrm{ml}$ Molarity $=\frac{120 / 60}{1120 / 1.15}=\frac{2 \times 1.15}{1120} \times 1000=2.05 \mathrm{M}$

Q. A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be (1) $\mathrm{MCl}_{2}$ (2) $\mathrm{MCl}_{4}$ (3) $\mathrm{MCl}_{5}$ (4) $\mathrm{MCl}_{3}$ [AIEEE 2012 (Online)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Mol. wt = 189.6 mass of $\mathrm{C} \ell=\frac{74.75}{100} \times 189.6=141.7 \mathrm{gm}$ formula: $\mathrm{MCl}_{4}$

Q. The ratio of number of oxygen atoms (O) in 16.0g ozone $\left(\mathrm{O}_{3}\right)$, 28.0 g carbon monoxide (CO) and 16.0g oxygen $\left(\mathrm{O}_{2}\right)$ is : (Atomic mass : $\mathrm{C}=12, \mathrm{O}=16$ and Avogadro’s constant $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{mol}^{-1}$ ) (1) 3 : 1 : 1         (2) 1 : 1 : 2            (3) 3 : 1 : 2            (4) 1 : 1 : 1 [AIEEE 2012 (Online)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) $\frac{16}{48} \times 3 \times \mathrm{N}_{\mathrm{A}}: \frac{28}{28} \times 1 \times \mathrm{N}_{\mathrm{A}}: \frac{16}{32} \times 2 \times \mathrm{N}_{\mathrm{A}}$ 1 : 1 : 1

Q. When $\mathrm{CO}_{2}$ (g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 litre of $\mathrm{CO}_{2}$ (g) over red hot coke, the total volume of the gases increased to 700 mL. The composition of the gaseous mixture at STP is :- (1) $\mathrm{CO}_{2}=200 \mathrm{mL} ; \mathrm{CO}=500 \mathrm{mL}$ (2) $\mathrm{CO}_{2}=350 \mathrm{mL} ; \mathrm{CO}=350 \mathrm{mL}$ (3) $\mathrm{CO}_{2}=0.0 \mathrm{mL} ; \mathrm{CO}=700 \mathrm{mL}$ (4) $\mathrm{CO}_{2}=300 \mathrm{mL} ; \mathrm{CO}=400 \mathrm{mL}$ [AIEEE 2012 (Online)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. An open vessel at 300 K is heated till $\frac{2}{5}$ th of the air in it is expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated is :- (1) 750 K        (2) 400 K            (3) 500 K          (4) 1500K [AIEEE 2012 (Online)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The density of 3M solution of sodium chloride is 1.252 g $\mathrm{mL}^{-1}$. The molality of the solution will be (molar mass, NaCl = 58.5 g $\mathrm{mol}^{-1}$) (1) 2.18 m          (2) 3.00 m            (3) 2.60 m           (4) 2.79 m [JEE(Main-online)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) $\mathrm{m}=\frac{\mathrm{M} \times 1000}{1000 \mathrm{d}-\mathrm{MMw}}=\frac{3 \times 1000}{1000 \times 1.252-3 \times 58.5}$ $=\frac{3000}{1252-175.5}=\frac{3000}{1076.5}=2.79$

Q. 10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration ? (1) 0.57 M          (2) 5.7 M              (3) 11.4 M            (4) 1.14 M [JEE(Main-online)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{M}_{\mathrm{f}}=\frac{2 \times 10+0.5 \times 200}{210}$ $=\frac{20+100}{210}=\frac{120}{210}=.57$

Q. Number of atoms in the following samples of substances is the largest in : (1) 127.0g of iodine (2) 48.0g of magnesium (3) 71.0g of chlorine (4) 4.0g of hydrogen [JEE(Main) 2013 (Online)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is (1) $\mathrm{C}_{2} \mathrm{H}_{4}$ (2) $\mathrm{C}_{3} \mathrm{H}_{4}$ ( 3) $\mathrm{C}_{6} \mathrm{H}_{5}$ (4) $\mathrm{C}_{7} \mathrm{H}_{8}$ [JEE(Main)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_{2} \rightarrow \mathrm{x} \mathrm{CO}_{2}+\frac{\mathrm{y}}{2} \mathrm{H}_{2} \mathrm{O}$ a’ moles $\quad$ xa’ $\quad\left(\frac{\mathrm{y}}{2} \times \mathrm{a}\right)$ $\frac{\mathrm{x}}{\mathrm{y} / 2}=\frac{3.08 / 44}{0.72 / 18}=\mathrm{C}_{7} \mathrm{H}_{8}$

Q. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of $\frac{\mathrm{M}}{10}$ sulphuric acid. The unreacted acid required 20 mL of $\frac{\mathrm{M}}{10}$ sodium hydroxide for complete neutralizaton. The percentage of nitrogen in the compound is : (1) 3% (2) 5% (3) 6% (4) 10% [JEE(Main-online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The amount of $\mathrm{BaSO}_{4}$ formed upon mixing 100 mL of 20.8% $\mathrm{BaCl}_{2}$ solution with 50 mL of 9.8% $\mathrm{H}_{2} \mathrm{SO}_{4}$ solution will be : (Ba = l37, Cl = 35.5, S=32, H = l and O = 16) (1) 33.2 g           (2) 11.65 g            (3) 23.3 g             (4) 30.6 g [JEE(Main-online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is : (1) 1 : 8         (2) 3 : 16           (3) 1 : 4          (4) 7 : 32 [JEE(Main)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Given $\frac{\mathrm{W}_{\mathrm{O}_{2}}}{\mathrm{W}_{\mathrm{N}_{2}}}=\frac{1}{4} \Rightarrow \frac{\mathrm{n}_{\mathrm{O}_{2}}}{\mathrm{n}_{\mathrm{N}_{2}}}=\frac{\mathrm{W}_{\mathrm{O}_{2}} \times \mathrm{M}_{\mathrm{N}_{2}}}{\mathrm{W}_{\mathrm{N}_{2}} \times \mathrm{M}_{\mathrm{O}_{2}}}=\frac{1}{4} \times \frac{28}{32}=\frac{7}{32}$

Q. The molecular formula of a commercial resin used for exchanging ions in water softening is $\mathrm{C}_{8} \mathrm{H}_{7} \mathrm{SO}_{3} \mathrm{Na}$ (Mol. w.t 206). What would be the maximum uptake of $\mathrm{Ca}^{2+}$ ions by the resin when expressed in mole per gram resin ? (1) $\frac{2}{309}$         (2) $\frac{1}{412}$           (3) $\frac{1}{103}$             (4) $\frac{1}{206}$ [JEE(Main)-2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) For softening of hard water by ion exchange resin method, reaction involved is

Q. 3g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is : (1) 42 mg         (2) 54 mg          (3) 18 mg          (4) 36 mg [JEE(Main)-2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\because$ Number of moles of $\mathrm{CH}_{3} \mathrm{COOH}$ adsorbed $=(0.06-0.042) \times \frac{50}{1000}$ $\therefore$ Amount of $\mathrm{CH}_{3} \mathrm{COOH}$ adsorbed per gram of charcoal $=\frac{0.018 \times 50}{1000} \times \frac{60}{3}=0.018 \mathrm{gm}$ = 18 mg

Q. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is : (at. mass Ag = 108; Br = 80) (1) 48        (2) 60        (3) 24           (4) 36 [JEE(Main)-2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% $\mathrm{O}_{2}$ by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :- (1) $\mathrm{C}_{4} \mathrm{H}_{10}$ (2) $\mathrm{C}_{3} \mathrm{H}_{6}$ (3) $\mathrm{C}_{3} \mathrm{H}_{8}$ (4) $\mathrm{C}_{4} \mathrm{H}_{8}$ [JEE(Main)-2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) If further information (i.e., 330 ml) is neglected, option (3) only satisfy the above equation.

Q. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of :- (1) Iron             (2) Fluoride              (3) Lead             (4) Nitrate [JEE Main-2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%) ; Carbon (22.9%), Hydrogen (10.0%) ; and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by $^{2} \mathrm{H}$ atoms is (1) 15 kg             (2) 37.5 kg              (3) 7.5 kg              (4) 10 kg [JEE(Main)-2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) Mass in the body of a healthy human adult has :- Oxygen = 61.4%, Carbon = 22.9%, Hydrogen = 10.0% and Nitrogen = 2.6% Total weight of person = 75 kg Mass due to $1 \mathrm{H}$ is $=75 \times \frac{10}{100}=7.5 \mathrm{kg}$ $^{1} \mathrm{H}$ atoms are replaced by $^{2} \mathrm{H}$ atoms. So mass gain by person =7.5 kg

Q. 1 gram of a carbonate $\left(\mathrm{M}_{2} \mathrm{CO}_{3}\right)$ on treatment with excess HCl produces 0.01186 mole of $\mathrm{CO}_{2}$. the molar mass of $\mathrm{M}_{2} \mathrm{CO}_{3}$ in g $\mathrm{mol}^{-1}$ is :- (1) 1186            (2) 84.3             (3) 118.6             (4) 11.86 [JEE(Main)-2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Given chemical eq $^{\mathrm{n}}$ $\mathrm{M}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl} \longrightarrow 2 \mathrm{MCl}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$ 1gm 0.01186mol $\Rightarrow$ from the balanced chemical $\mathrm{eq}^{\mathrm{n}}$ $\frac{1}{\mathrm{M}}=0.01186$ .

Q. A water sample has ppm level concentration of following anions $\mathrm{F}^{-}=10 ; \mathrm{SO}_{4}^{2-}=100 ; \mathrm{NO}_{3}^{-}=50$ the anion/anions that make / makes the water sample unsuitable for drinking is / are : (1) only $\mathrm{NO}_{3}^{-}$ (2) both $\mathrm{SO}_{4}^{2-}$ and $\mathrm{NO}_{3}^{-}$ (3) only $\mathrm{F}^{-}$ (4) only $\mathrm{SO}_{4}^{2-}$ [JEE – Main 2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\mathrm{NO}_{3}^{-}$ : The maximum limit of nitrate in drinking water is 50 ppm. Excess nitrate in drinking water can cause disease. Such as methemoglobinemia. $\mathrm{SO}_{4}^{2-}:$ above 500 ppm of $\mathrm{SO}_{4}^{2-}$ ion in drinking water causes laxative effect otherwise at moderate levels it is harmless $\mathrm{F}^{-}$ : Above 2ppm concentration of $\mathrm{F}^{-}$ in drinking water cause brown mottling of teeth. The concentration given in question of $\mathrm{SO}_{4}^{2-} \& \mathrm{NO}_{3}^{-}$ in water is suitable for drinking but the concentration of $\mathrm{F}^{-}$ (i.e 10 ppm) make water unsuitable for drinking purp

Q. The ratio of mass percent of C and H of an organic compound $\left(\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}\right)$ is 6 : 1. If one molecule of the above compound $\left(\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}\right)$ contains half as much oxygen as required to burn one molecule of compound $\mathrm{C}_{\mathrm{X}} \mathrm{H}_{\mathrm{Y}}$ completely to $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$. The empirical formula of compound $\mathrm{C}_{\mathrm{X}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}$ is : (1) $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}$ (2) $\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}$ (3) $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{3}$ (4) $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$ [JEE(Main)-2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Ionic Equilibrium – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.   Simulator   Previous Years AIEEE/JEE Mains Questions
Q. Solid Ba$\left(\mathrm{NO}_{3}\right)_{2}$ is gradully dissolved in a 1.0 × $10^{-4} \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ solution.At what concentration of Ba2+ will a precipitate begin to form? $\left(\mathrm{K}_{\mathrm{SP}} \text { for } \mathrm{Ba} \mathrm{CO}_{3}=5.1 \times 10^{-9}\right)$ (A) $8.1 \times 10^{-8} \mathrm{M}$ (B) $8.1 \times 10^{-7} \mathrm{M}$ (C) $4.1 \times 10^{-5} \mathrm{M}$ (D) $5.1 \times 10^{-5} \mathrm{M}$ [AIEEE-2009,JEE-MAIN(Online)–2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D) $5.1 \times 10^{-9}=\left[\mathrm{Ba}^{+2}\right]\left[10^{-4}\right]$ $\left[\mathrm{Ba}^{+2}\right]=5.1 \times 10^{-5} \mathrm{M}$

Q. At 25° C, the solubility producct of $\mathrm{Mg}(\mathrm{OH})_{2}$ is $1.0 \times 10^{-11}$. At which pH, will $\mathrm{Mg}^{2+}$ ions start precipitating in the form of $\mathrm{Mg}(\mathrm{OH})_{2}$ from a solution of 0.001 M $\mathrm{Mg}^{2+}$ ions? (A) 8                     (B) 9                        (C) 10                             (D) 11 [AIEEE–2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C) $10^{-11}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{OH}^{-}\right]^{2}$ $10^{-11}=\left(10^{-3}\right)\left[\mathrm{OH}^{-}\right]^{2}$ $\left[\mathrm{OH}^{-}\right]=10^{-4} \quad \mathrm{pOH}=4 \quad \mathrm{pH}=11$

Q. In aqueous solution the ionization constants for carbonic acid are $\mathrm{K}_{1}=4.2 \times 10^{-7}$ and $\mathrm{K}_{2}=4.8$ $\times 10^{-11}$ Select the correct statement for a saturated 0.034 M solution of the carbonic acid :- (A) The concentration of $\mathrm{H}^{+}$ is double that of $\mathrm{CO}_{3}^{2-}$ (B) The concentration of $\mathrm{CO}_{3}^{2-}$ is $0.034 \mathrm{M}$ (C) The concentration of $\mathrm{CO}_{3}^{2-}$ is greater than that of $\mathrm{HCO}_{3}^{-}$ (D) The concentrations of $\mathrm{H}^{+}$ and $\mathrm{HCO}_{3}^{-}$ are approximately equal [AIEEE–2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D)

Q. Solubility product of silver bromide is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass taken as 120 g $\left.\mathrm{mol}^{-1}\right)$ to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is :- (A) $5.0 \times 10^{-8} \mathrm{g}$ (B) $1.2 \times 10^{-10} \mathrm{g}$ (C) $1.2 \times 10^{-9} \mathrm{g}$ (D) $6.2 \times 10^{-5} \mathrm{g}$ [AIEEE–2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C) $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]=\mathrm{Ksp}$ $[0.05]\left[\frac{\mathrm{W}}{120}\right]=5 \times 10^{-13}$ $\mathrm{w}=\frac{120 \times 5 \times 10^{-13}}{5 \times 10^{-2}}=120 \times 10^{-11}=12 \times 10^{-10}$ $=1.2 \times 10^{-9} \mathrm{g}$

Q. An acid HA ionises as The pH of 1.0 M solution is 5. Its dissociation constant would be :- (A) $1 \times 10^{-10}$ (B) 5 (C) $5 \times 10^{-8}$ (D) $1 \times 10^{-5}$ [AIEEE–2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A) $\left[\mathrm{H}^{+}\right]=10^{-5}=\mathrm{C}_{\mathrm{o}} \alpha$ $\alpha=10^{-5}$

Q. The $\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{Cr}(\mathrm{OH})_{3}$ is $1.6 \times 10^{-30}$ The molar solubility of this compound in water is :- (A) $\sqrt[2]{1.6 \times 10^{-30}}$ (B) $\sqrt[4]{1.6 \times 10^{-30}}$ (C) $\sqrt[4]{1.6 \times 10^{-30} / 27}$ (D) $1.6 \times 10^{-30 / 27}$ [AIEEE–2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C) $\mathrm{Ksp}=1.6 \times 10^{-30}=27 \mathrm{S}^{4}$ $S^{4}=\left[\frac{1.6 \times 10^{-30}}{27}\right]$ $S=\left[\frac{1.6 \times 10^{-30}}{27}\right]^{1 / 4}$

Q. The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is :- (A) $1 \times 10^{-7}$ (B) $3 \times 10^{-7}$ (C) $1 \times 10^{-3}$ (D) $1 \times 10^{-5}$ [AIEEE–2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D) [\mathrm{HQ}]=0.10 \mathrm{M} \mathrm{pH}=3 \quad ;\left[\mathrm{H}^{+}\right]=10^{-3}=\mathrm{C}_{0} \alpha \alpha=10^{-2}

Q. If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CaF}_{2}$ at $25^{\circ} \mathrm{C}$ is $1.7 \times 10^{-10}$ , the combination amongst the following which gives a precipitate of $\mathrm{CaF}_{2}$ is :- (A) $1 \times 10^{-2} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-5} \mathrm{M} \mathrm{F}^{-}$ (B) $1 \times 10^{-4} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-4} \mathrm{M} \mathrm{F}^{-}$ (C) $1 \times 10^{-3} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-5} \mathrm{M} \mathrm{F}^{-}$ (D) $1 \times 10^{-2} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-3} \mathrm{M} \mathrm{F}^{-}$ [JEE-MAIN(online)–2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D)

Q. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ? (A) 0.1 L             (B) 0.9 L              (C) 2.0 L            (D) 9.0 L [AIEEE–2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D) \left(10^{-1}\right)(1)=\left(10^{-2}\right)(1+\mathrm{v}) 10=\mathrm{v}+1 v = 9L

Q. What would be the pH of a solution obtained by mixing 5 g of acetic acid and 7.5 g of sodium acetate and making the volume equal to 500 mL? $\left(\mathrm{Ka}=1.75 \times 10^{-5}, \mathrm{pKa}=4.76\right)$ (A) 4.76 < pH < 5.0 (B) pH < 4.70 (C) pH of solution will be equal to pH of acetic acid (D) pH = 4.70 [JEE-MAIN(Online)–2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)

Q. Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{BaSO}_{4}$ and $\mathrm{CrCl}_{3}$ respectively ? (A) $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{\frac{1}{3}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{\frac{1}{3}},\left(\mathrm{K}_{\mathrm{sp}}\right)^{\frac{1}{2}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{\frac{1}{4}}$ (B) $\left(\mathrm{K}_{\mathrm{ap}}\right)^{\frac{1}{2}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{\frac{1}{3}},\left(\frac{\mathrm{K}_{\mathrm{gp}}}{27}\right)^{\frac{1}{4}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{\frac{1}{3}}$ (C) $\left(\mathrm{K}_{\mathrm{sp}}\right)^{\frac{1}{2}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{\frac{1}{3}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{\frac{1}{4}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{\frac{1}{3}}$ $(\mathrm{D})\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{\frac{1}{3}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{\frac{1}{4}},\left(\mathrm{K}_{\mathrm{sp}}\right)^{\frac{1}{2}},\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{\frac{1}{3}}$ [JEE-MAIN(Online)–2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)

Q. NaOH is a strong base. What will be pH of 5.0 × $10^{-2} \mathrm{M}$ NaOH solution ? (log2 = 0.3) (A) 13.70             (B) 13.00             (C) 14.00             (D) 12.70 [JEE-MAIN(Online)–2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D)

Q. Zirconium phosphate $\left[\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}\right)_{4}\right]$ dissociates into three zirconium cations of charge +4 and four phosphate anions of charge –3. If molar solubility of zirconium phosphate is denoted by S and its solubility product by $\mathrm{K}_{\mathrm{sp}}$ then which of the following relationship between S and $\mathrm{K}_{\mathrm{sp}}$is correct ? (A) $\mathrm{S}=\left\{\mathrm{K}_{\mathrm{sp}} / 144\right\}^{1 / 7}$ (B) $\mathrm{S}=\left\{\mathrm{K}_{\mathrm{sp}} /(6912)^{1 / 7}\right\}$ (C) $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 6912\right)^{1 / 7}$ (D) $\mathrm{S}=\left\{\mathrm{K}_{\mathrm{sp}} / 6912\right\}^{7}$ [JEE-MAIN(Online)–2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C)

Q. In some solutions, the concentration of $\mathrm{H}_{3} \mathrm{O}^{+}$ remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as :- (A) Colloidal solutions (B) True solutions (C) Ideal solutions (D) Buffer solutions [JEE-MAIN(Online)–2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D)

Q. An aqueous solution contains 0.10 M $\mathrm{H}_{2} \mathrm{S}$ and 0.20 M HCl. If the equilibrium constants for the formation of HS– from H2S is 1.0 × $10^{-7}$ and that of $\mathrm{S}^{2-}$ from $\mathrm{HS}^{-}$ ions is 1.2×$10^{-13}$ then the concentration of $\mathrm{S}^{2-}$ ions in aqueous solution is : (A) $3 \times 10^{-20}$ (B) $6 \times 10^{-21}$ (C) $5 \times 10^{-19}$ (D) $5 \times 10^{-8}$ [JEE-MAIN–2018]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)

Chemical Equilibrium – JEE Mains Previous Year Questions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.   Simulator   Previous Years AIEEE/JEE Mains Questions
Q. A vessel at 1000 K contains $\mathrm{CO}_{2}$ with a pressure of 0.5 atm. Some of the $\mathrm{CO}_{2}$ is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :- (1) 0.3 atm (2) 0.18 atm (3) 1.8 atm (4) 3 atm [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The equilibrium constant $\left(\mathrm{K}_{\mathrm{C}}\right)$ for the reaction $\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})$ at temperature T is $4 \times 10^{-4}$ The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction. $\mathrm{NO}(\mathrm{g}) \longrightarrow 1 / 2 \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$ at the same temperature is :- (1) 50.0 (2) 0.02 (3) $2.5 \times 10^{2}$ (4) $4 \times 10^{-4}$ [AIEEE-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{K}_{\mathrm{c}}=\frac{1}{\sqrt{\mathrm{K}_{\mathrm{c}}}}$

Q. 8 mol of $\mathrm{AB}_{3}(\mathrm{g})$ are introduced into a 1.0 $\mathrm{d} \mathrm{m}^{3}$ vessel. If it dissociates as $2 \mathrm{AB}_{3}(\mathrm{g}) \square \quad \mathrm{A}_{2}(\mathrm{g})+3 \mathrm{B}_{2}(\mathrm{g})$ At equilibrium, 2mol of $\mathrm{A}_{2}$ are found to be present. The equilibrium constant of this reaction is :- (1) 36 (2) 3 (3) 27 (4) 2 [JEE-MAINS(online)-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. The value of Kp for the equilibrium reaction $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \square 2 \mathrm{NO}_{2}(\mathrm{g})$ is 2 The percentage dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})$ at a pressure of 0.5 atm is (1) 71 (2) 50 (3) 88 (4) 25 [JEE-MAINS(online)-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{K}_{\mathrm{p}}=\frac{(2 \alpha)^{2}}{(1-\alpha)} \times \frac{0.5}{(1+\alpha)}$

Q. $\mathrm{K}_{1}, \mathrm{K}_{2}$ and $\mathrm{K}_{3}$ are the equilibrium constants of the following reactions (I), (II) and (III), respectively (I) $\mathrm{N}_{2}+2 \mathrm{O}_{2} \square 2 \mathrm{NO}_{2}$ (II) $2 \mathrm{NO}_{2} \square \mathrm{N}_{2}+2 \mathrm{O}_{2}$ (III) $\mathrm{NO}_{2} \square \frac{1}{2} \mathrm{N}_{2}+\mathrm{O}_{2}$ The correct relation from the following is : (1) $\mathrm{K}_{1}=\sqrt{\mathrm{K}_{2}}=\mathrm{K}_{3}$ (2) $\mathrm{K}_{1}=\frac{1}{\mathrm{K}_{2}}=\frac{1}{\mathrm{K}_{3}}$ $(3) \mathrm{K}_{1}=\frac{1}{\mathrm{K}_{2}}=\mathrm{K}_{3}$ (4)$\mathrm{K}_{1}=\frac{1}{\mathrm{K}_{2}}=\frac{1}{\left(\mathrm{K}_{3}\right)^{2}}$ [JEE-MAINS(online)-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Fact

Q. One mole of $\mathrm{O}_{2}(\mathrm{g})$ and two moles of SO2(g) were heated in a closed vessel of one litre capacity at 1098 K. At equilibrium 1.6 moles of $\mathrm{SO}_{3}$ (g) were found. The equilibrium constant $\mathbf{K}_{C}$ of the reaction would be :- (1) 60 (2) 80 (3) 30 (4) 40 [JEE-MAINS(online)-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. $\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \square 2 \mathrm{NH}_{3}(\mathrm{g}), \mathrm{K}_{1}$ $\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \square 2 \mathrm{NO}(\mathrm{g}), \mathrm{K}_{2} \quad(\mathrm{B})$ $\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \square \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{K}_{3} \quad(\mathrm{C})$ The equation for the equilibrium constant of the reaction $2 \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{g}) \square 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\left(\mathrm{K}_{4}\right)$ in terms of $\mathrm{K}_{1}, \mathrm{K}_{2}$ and $\mathrm{K}_{3}$ is : (1) $\frac{\mathrm{K}_{1} \mathrm{K}_{3}^{2}}{\mathrm{K}_{2}}$ (2) $\frac{\mathrm{K}_{2} \mathrm{K}_{3}^{3}}{\mathrm{K}_{1}}$ (3) $\frac{\mathrm{K}_{1} \mathrm{K}_{2}}{\mathrm{K}_{3}}$ (4) $\mathrm{K}_{1} \mathrm{K}_{2} \mathrm{K}_{3}$ [JEE-MAINS(online)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Fact

Q. In reaction $\mathrm{A}+2 \mathrm{B} \square 2 \mathrm{C}+\mathrm{D}$, initial concentration of B was 1.5 times of |A|, but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is (1)4 (2) 6 (3) 12 (4) 8 [JEE-MAINS(online)-2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{K}_{\mathrm{c}}=4$

Q. For the decomposition of the compound, represented as If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be (1) $38.8 \times 10^{-2}$ atm (2) $1.94 \times 10^{-2}$ atm (3) $5.82 \times 10^{-2}$ atm (4) $7.66 \times 10^{-2}$ atm [JEE-MAINS(online)-2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. For the reaction $\mathrm{SO}_{2(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{g})} \square \mathrm{SO}_{3(\mathrm{g})},$ if $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\mathrm{x}}$ where the symbols have usual meaning then the value of x is : (assuming ideality) ( 1)$\frac{1}{2}$ ( 2) 1 (3) –1 $(4)-\frac{1}{2}$ [JEE-MAINS 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{-\frac{1}{2}}$

Q. The equilibrium constants at 298 K for a reaction $\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$ is 100 If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol $\mathrm{L}^{-1}$) will be : (1) 1.182 (2) 0.182 (3) 0.818 (4) 1.818 [JEE-Mains 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Atomic Structure – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.   Simulator   Previous Years AIEEE/JEE Mains Questions
Q. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is ($\mathrm{Ch}=6.6 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}$, mass of electron, $\mathrm{e}_{\mathrm{m}}=9.1 \times 10^{-31} \mathrm{kg}$):- (1) $1.92 \times 10^{-3} \mathrm{m}$ (2) $3.84 \times 10^{-3} \mathrm{m}$ (3) $1.52 \times 10^{-4} \mathrm{m}$ (4) $5.10 \times 10^{-3} \mathrm{m}$ [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. Calculate the wavelength (in nanometer) associated with a proton moving at $1.0 \times 10^{3} \mathrm{ms}^{-1}$ (Mass of proton = $1.67 \times 10^{-27} \mathrm{kg}$ and $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$) :- (1) 2.5 nm (2) 14.0 nm (3) 0.032 nm (4) 0.40 nm [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) $\mathrm{m}_{\mathrm{p}}=1.67 \times 10^{-27}$ $\mathrm{h}=6.63 \times 10^{-34}$ $\mathrm{v}=10^{3}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 10^{3}}$ $=3.97 \times 10^{-7+3}$ $=3.97 \times 10^{-10}$ $=\frac{3.9 \times 10^{-10}}{10^{-9}} \mathrm{nm} \quad=0.40 \mathrm{nm}$

Q. The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ $\mathrm{mol}^{-1}$. The longest wavelength of light capable of breaking a single Cl–Cl bond is $\left(\mathrm{C}=3 \times 10^{8} \mathrm{ms}^{-1} \text { and } \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{mol}^{-1}\right)$ (1) 494 nm (2) 594 nm (3) 640 nm (4) 700 nm [AIEEE-2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{B.E.}=242 \mathrm{kJ} / \mathrm{mol}$ $\mathrm{E}=\frac{\mathrm{hcN}_{\mathrm{A}}}{\lambda}$ $10^{3} \times 242 \times \lambda=3 \times 10^{8} \times 6.626 \times 10^{-34} \times 6.02 \times 10^{23}$ $\lambda=\frac{3 \times 6.626 \times 6.02 \times 10^{-26+23}}{242}$ $=0.494 \times 10^{-3} \times 10^{-3}$ = 494 nm

Q. Ionisation energy of $\mathrm{He}^{+}$ is $19.6 \times 10^{-18} \mathrm{J}$ atom $^{-1}$. The energy of the first stationary state (n = 1) of $\mathrm{L} \mathbf{i}^{2+}$ is:- (1) $8.82 \times 10^{-17} \mathrm{J}$ atom $^{-1}$ (2) $4.41 \times 10^{-16} \mathrm{J}$ atom $^{-1}$ (3) $-4.41 \times 10^{-17} \mathrm{J}$ atom $^{-1}$ (4) $-2.2 \times 10^{-15} \mathrm{J}$ atom $^{-1}$ [AIEEE-2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) I.E. $=19.6 \times 10^{-18}$ I.E $\propto \mathrm{z}^{2}$ $\frac{(\mathrm{I.E.})_{\mathrm{Li}^{+2}}}{(\mathrm{I.E.})_{\mathrm{He}}}=\frac{\mathrm{Z}_{\mathrm{Li}}^{2}}{\mathrm{Z}_{\mathrm{He}}^{2}} \quad \mathrm{E}_{1}=\frac{9}{4} \times 19.6 \times 10^{-18}$ $=-4.41 \times 10^{-17}$

Q. The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following (1) n = 3 to n = 1 (2) n = 2 to n = 1 (3) n = 3 to n = 2 (4) n = 4 to n = 3 [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. The electrons identified by quantum numbers n and  :- (a) n = 4 ,  = 1 (b) n = 4,  = 0 (c) n = 3,  = 2 (d) n = 3,  = 1 Can be placed in order of increasing energy as (1) (a) < (c) < (b) < (d) (2) (c) < (d) < (b) < (a) (3) (d) < (b) < (c) < (a) (4) (b) < (d) < (a) < (c) (3) (d) < (b) < (c) < (a) (4) (b) < (d) < (a) < (c) [AIEEE-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) (d) < (b) < (c) < (a) Acc. to (n + ) rule.

Q. If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become :- (1) Two times (2) Half (3) One fourth (4) Four time [JEE-Main(online2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) $\lambda \propto \frac{1}{\sqrt{\mathrm{KE}}}$

Q. The wave number of the first emission line in the Balmer series of H-Spectrum is : (R = Rydberg constant) : (1) $\frac{3}{4} \mathrm{R}$ (2) $\frac{9}{400} \mathrm{R}$ (3) $\frac{5}{36} \mathrm{R}$ (4) $\frac{7}{6} \mathrm{R}$ [JEE-Main(online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\bar{v}=\frac{1}{\mathrm{R}}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5}{36 \mathrm{R}}$

Q. The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is : $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{J} \mathrm{s}\right)$ (1) $6.626 \times 10^{-31} \mathrm{m}$ (2) $6.626 \times 10^{-34} \mathrm{m}$ (3) $6.626 \times 10^{-38} \mathrm{m}$ (4) $6.626 \times 10^{-30} \mathrm{m}$ [JEE-Main(online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship? (1) a dust particle (2) an electron (3) a proton (4) an -particle. [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of $\mathbf{L} \mathbf{i}^{++}$ is : (1) 13.6 eV (2) 30.6 eV (3) 122.4 eV (4) 3.4 eV [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) B.E. $=3.4 \times 9=30.6 \mathrm{eV}$

Q. Based on the equation $\Delta \mathrm{E}=-2.0 \times 10^{-18} \mathrm{J}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ the wavelength of the light that must be absorbed to excite hydrogen electron from level n = 1 to level n $=2$ will be $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{Js}, \mathrm{C}=3 \times 10^{8} \mathrm{ms}^{-1}\right)$ (1) $2.650 \times 10^{-7} \mathrm{m}$ (2) $1.325 \times 10^{-7} \mathrm{m}$ (3) $1.325 \times 10^{-10} \mathrm{m}$ (4) $5.300 \times 10^{-10} \mathrm{m}$ [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) $\frac{1}{\lambda}=\frac{2 \times 10^{-18}}{\mathrm{hc}}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $\Rightarrow \frac{1}{\lambda}=\frac{2 \times 10^{-18}}{6.625 \times 10^{-34} \times 3 \times 10^{8}} \times \frac{3}{4}$ $\Rightarrow \lambda=\frac{2 \times 6.625 \times 10^{-34} \times 10^{8}}{10^{-18}}$ $=13.25 \times 10^{-8}$ $=1.325 \times 10^{-7} \mathrm{m}$

Q. If be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{E}=\phi+\frac{1}{2} \mathrm{mv}^{2}$ $\Rightarrow \frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\frac{1}{2} \mathrm{mv}^{2}$ $\Rightarrow \mathrm{v}^{2}=\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right] \Rightarrow \mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]}$ $\Rightarrow \mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right]}$

Q. Ionization energy of gaseous Na atoms is 495.5 $\mathrm{kjmol}^{-1}$ . The lowest possible frequency of light that ionizes a sodium atom is $\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{mol}^{-1}\right)$ (1) $3.15 \times 10^{15} \mathrm{s}^{-1}$ (2) $4.76 \times 10^{14} \mathrm{s}^{-1}$ (3) $1.24 \times 10^{15} \mathrm{s}^{-1}$ (4) $7.50 \times 10^{4} \mathrm{s}^{-1}$ [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\Delta \mathrm{E}=\mathrm{hv}$ $\mathrm{v}=\frac{\Delta \mathrm{E}}{\mathrm{h}}$ $\mathrm{v}=\frac{495.5 \times 10^{3} \mathrm{Joule}}{6.023 \times 10^{23}} \times \frac{1}{6.626 \times 10^{-34}}$ $\mathrm{v}=1.24 \times 10^{15} \mathrm{sec}^{-1}$

Q. Which of the following is the energy of a possible excited state of hydrogen? (1) –3.4 eV (2) +6.8 eV (3) +13.6 eV (4) –6.8 eV [JEE-Main(offline) 2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) For H-atom, (Z = 1) $\mathrm{E}_{\mathrm{n}}=-13.6 \times \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}$ $\begin{aligned} \therefore \text { So for } \mathrm{E}_{1} &=-13.6 \mathrm{eV} \\ \mathrm{E}_{2} &=-3.4 \mathrm{eV} \end{aligned}$

Q. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron respectively, then the value of $\mathrm{h} / \lambda$ (where $\lambda$is wavelength associated with electron wave) is given by : (1) $\sqrt{2 \mathrm{meV}}$ (2) mev (3) $2 \mathrm{meV}$ (4) $\sqrt{\mathrm{meV}}$ [JEE-Main 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) As electron of charge ‘e’ is passed through ‘V’ volt, kinetic energy of electron becomes = ‘eV’ As wavelength of $e^{-}$ wave = $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m} \cdot \mathrm{K} \cdot \mathrm{E}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\therefore \quad \frac{\mathrm{h}}{\lambda}=\sqrt{2 \mathrm{meV}}$

Q. The radius of the second Bohr orbit for hydrogen atom is : (Planks const. $\mathrm{h}=6.6262 \times 10^{-34} \mathrm{Js}$; mass of electron $=9.1091 \times 10^{-31} \mathrm{kg} ;$ charge of electron $\mathrm{e}=$ $1.60210 \times 10^{-19} \mathrm{C}:$ permittivity of vaccuml $\left.\epsilon_{0}=8.854185 \times 10^{-12} \mathrm{kg}^{-1} \mathrm{m}^{-3} \mathrm{A}^{2}\right)$ [JEE-Main 2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Radius of $\mathbf{n}^{\mathrm{th}}$ Bohr orbit in H-atom Radius of II Bohr orbit $=0.53 \times(2)^{2}$

Structural Isomerism – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.   Simulator Previous Years AIEEE/JEE Mains Questions
Q. The alkene that exhibits geometrical isomerism is :- (1) 2–butene (2) 2–methyl–2–butene (3) Propene (4) 2–methyl propene [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. The number of stereoisomers possible for a compound of the molecular formula CH3–CH=CH–CH(OH)–Me is :- (1) 4             (2) 6              (3) 3             (4) 2 [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) exhibits both geometrical as well as optical isomerism. Cis – R Trans – R Cis – S Trans – S

Q. Out of the following, the alkene that exhibits optical isomerism is : (1) 2-methyl-2-pentene (2) 3-methyl-2-pentene (3) 4-methyl-1-pentene (4) 3-methyl-1-pentene [AIEEE-2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Identify the compound that exhibits tautomerism :- (1) 2-Pentanone (2) Phenol (3) 2-Butene (4) Lactic acid [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. The IUPAC name of the following compounds is : (1) (Z) – 5 – hepten – 3 – yne (2) (Z) – 2 – hepten – 4 – yne (3) (E) – 5 – hepten – 3 – yne (4) (E) – 2 – hepten – 4 – yne [JEE-Main-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

Q. Dipole moment is shown by :- (1) trans-2, 3-dichloro- 2-butene (2) 1, 2-dichlorobenzene (3) 1, 4-dichlorobenzene (4) trans-1, 2-dinitroethene [JEE-Main 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)

Q. Maleic acid and fumaric acids are :- (1) Tautomers (2) Chain isomers (5) Geometrical isomers (4) Functional isomers [JEE-Main 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. Monocarboxylic acids are functional isomers of : (1) Esters (2) Amines (3) Ethers (4) Alcohols [JEE-Main 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) Monocarboxylic acid and Esters $(-\mathrm{COO}-)$ has same general $(-\mathrm{COOH}) .$ Formula $\left[\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}} \mathrm{O}_{2}\right]$ but different FG, so are called functional group isomer.

Q. Arrange in the correct order of stability (decreasing order) for the following molecules: [JEE-Main 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. For which of the following molecule significant $\mu \neq 0$ [JEE-Main 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) Due to presence of $\ell . \mathrm{p}_{(\mathrm{s})}$ on oxygen and sulphur atom which are out of plane hence

 
Oxidation & Reduction – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.   Simulator   Previous Years AIEEE/JEE Main Questions
Q. Which of the following will change the colour of acidic dichromate solution. [JEE-MAINS 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)

Q. [JEE-MAINS 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C)

Q. [JEE-MAINS 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)

Q. The most suitable reagent for the conversion of $\mathrm{R}-\mathrm{CH}_{2}-\mathrm{OH} \rightarrow \mathrm{R}-\mathrm{CHO}$ is :- (A) $\mathrm{CrO}_{3}$ (B) PCC (Pyridinium chlorochromate) (D) $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ [JEE-MAINS 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (B)