Mind Maps for Rotational Motion: Torque Revision Class XI, JEE, NEET

Rotational Motion Class 11 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the key formulae and important concepts on finger tips.

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Rotational Motion part2

 


 

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NTA NEET 2020 Application Form (Open Again) | Admit Card, Exam Date, Pattern, Syllabus & Preparation

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Mind Maps for Electromagnetic Waves Revision – Class XII, JEE, NEET

Get to learn all the formulae and important points of Electromagnetic Waves through this mind map. Download and share with your friends also.

 

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Mind Maps for Significant Figures-Error In Measurement Revision – Class XII, JEE, NEET

Learn all the concepts of Significant Figures through this mind map.

 

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Significant Figures


 

Mind Maps for Vectors Revision – Class XI, JEE, NEET

Learn all the formulas and important topics of Vectors through this mind map.

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Revise Complete Physics for Class 11, 12, JEE & NEET. Click Here 


 

Mind Map for Modern Physics: Radioactivity Revision – Class XII, JEE, NEET

Radioactivity in Modern Physics comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

 

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Radioactivity Mind Map

 

 


 

Mind Maps for Modern Physics: Nuclei Revision – Class XII, JEE, NEET

Nuclear Physics in Class 12 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

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Nuclear Physics mind Map

 


 

Mind Maps for Photo Electric Effect Revision – Class XII, JEE, NEET

Photoelectric Effect in Class 12 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

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Photoelectric effect

 

 


 

Wave Optics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Previous Years JEE Advanced Questions

Q. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits $S_{1}$ and $S_{2}$. In each of these cases $S_{1} P_{0}$ = $S_{2} P_{0}$, $S_{1} P_{1}$$S_{2} P_{1}$ = $\lambda / 4$ and $S_{1} P_{2}$$S_{2} P_{2}=\lambda / 3$, where $\lambda$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $\mu$ and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by $\delta$ (P) and the intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation.

[IIT-JEE-2009]

Sol. ((A) $p, s ;(B) q ;(C) t ;(D) r, s, t$)

(A) $\Delta \mathrm{x}=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=0$

$\delta\left(\mathrm{P}_{0}\right)=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=0$

$\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}=\frac{\lambda}{4}$

$\delta\left(\mathrm{P}_{1}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$

$\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\Delta \phi}{2}\right)$

$\mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{1}=\mathrm{I}_{\max } \cos ^{2} \frac{\delta}{2}=\frac{\mathrm{I}_{\max }}{2}$

$\delta\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}$

$\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{2}=\mathrm{I}_{\max } \cos ^{2} \frac{\pi}{3}=\frac{\mathrm{I}_{\max }}{4}$

$\mathrm{I}\left(\mathrm{P}_{0}\right)>\mathrm{I}\left(\mathrm{P}_{1}\right)$

$(\mathrm{B}) \Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}-\left[\mathrm{S}_{2} \mathrm{P}+(\mu-1) \mathrm{t}\right]$

$\Delta \mathrm{x}_{1}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}-(\mu-1) \mathrm{t}$

$\Delta \mathrm{x}_{1}=\frac{\lambda}{4}-\frac{\lambda}{4}=0$

$8\left(\mathrm{P}_{1}\right)=0 ; \mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{\max }$

$8\left(\mathrm{P}_{0}\right)=\frac{\pi}{2} \delta\left(\mathrm{P}_{0}\right) \neq 0$

$\mathrm{I}\left(\mathrm{P}_{0}\right)=\mathrm{I}_{\max } / 2$

$\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{2}-\mathrm{S}_{1} \mathrm{P}_{2}-(\mu-1) \mathrm{t}$

$=\frac{\lambda}{3}-\frac{\lambda}{4}=\frac{\lambda}{12}$

$8\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{12}=\frac{\pi}{6}$

$\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{12}\right)$


Q. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $\beta_{G}, \beta_{R}$ and $\beta_{B},$ respectively. Then

(A) $\beta_{G}>\beta_{B}>\beta_{R}$

(B) $\beta_{B}>\beta_{G}>\beta_{R}$

(C) $\beta_{R}>\beta_{B}>\beta_{G}$

(D) $\beta_{R}>\beta_{G}>\beta_{B}$

[IIT-JEE-2012]

Sol. (D)

$\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$

$\lambda_{\mathrm{R}}>\lambda_{\mathrm{a}}>\lambda_{\mathrm{B}}$


Q. In the Young’s double slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is :-

(A) $(2 n+1) \frac{\lambda}{2}$

(B) $(2 n+1) \frac{\lambda}{4}$

(C) $(2 n+1) \frac{\lambda}{8}$

$(D)(2 n+1) \frac{\lambda}{16}$

[JEE Advanced 2013]

Sol. (B)

$\frac{\mathrm{I}_{\max }}{2}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)$

$\cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\frac{1}{2}$

$\cos \left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\pm \frac{1}{\sqrt{2}}$

$\frac{\pi}{\lambda} \Delta \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{4}$

$\Delta \mathrm{x}=\left(\mathrm{n} \pm \frac{1}{4}\right) \lambda$


Q. A light source, which emits two wavelengths $\lambda_{1}=400 \mathrm{nm}$ and $\lambda_{2}=600 \mathrm{nm},$ is used in a Young’s double slit experiment. If recorded fringe widths for $\lambda_{1}$ and $\lambda_{2}$ are $\beta_{1}$ and $\beta_{2}$ and the number of fringes for them within a distance y on one side of the central maximum are $\mathrm{m}_{1}$ and $\mathrm{m}_{2},$ respectively, then :-

(A) $\beta_{2}>\beta_{1}$

(B) $\mathrm{m}_{1}>\mathrm{m}_{2}$

(C) From the central maximum, $3^{\mathrm{rd}}$ maximum of $\lambda_{2}$ overlaps with $5^{\text {th }}$ minimum of $\lambda_{1}$

(D) The angular separation of fringes of $\lambda_{1}$ is greater than $\lambda_{2}$

[JEE Advanced 2014]

Sol. (A,B,C)

$\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$

$\mathrm{B}_{2}>\beta_{1}$

$\mathrm{y}=\mathrm{m}_{1} \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}}=\mathrm{m}_{2} \frac{\mathrm{D} \lambda_{2}}{\mathrm{d}}$

$\frac{\mathrm{nD} \times \lambda_{2}}{\mathrm{d}}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}} \Rightarrow 600 \mathrm{n}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \times 4$


Q. A Young’s double slit interference arrangement with slits $S_{1}$ and $S_{2}$ is immersed in water (refractive index $=4 / 3$ ) as shown in the figure. The positions of maxima on the surface of water are given by $x^{2}=p^{2} m^{2} \lambda^{2}-d^{2},$ where $\lambda$ is the wavelength of light in air (refractive index $=1$, $2 d$ is the separation between the slits and $m$ is an integer. The value of p is.

[JEE Advanced 2015]

Sol. 3


Q. While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources $\left(\mathrm{S}_{1}, \mathrm{S}_{2}\right)$ emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3m from the mid-point of $\mathrm{S}_{1} \mathrm{S}_{2}$, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining $\mathrm{S}_{1} \mathrm{S}_{2}$. Which of the following is (are) true of the intensity pattern on the screen ?

(A) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction

(B) Semi circular bright and dark bands centered at point O

(C) The region very close to the point O will be dark

(D) Straight bright and dark bands parallel to the x-axis

[JEE-Mains 2016]

Sol. (B,C)

Path difference at point O = d = .6003 mm = 600300 nm

$=\frac{2001}{2}(600 \mathrm{nm})=1000 \lambda+\frac{\lambda}{2}$

$\Rightarrow$ minima form at point $\mathrm{O}$

Line $S_{1} S_{2}$ and screen are $\perp$ to each other so fringe pattern is circular (semi-circular because only half of screen is available)


Q. Two coherent monochromatic  point sources $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of wavelength $\lambda$ = 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta \theta$. Which of the following options is/are correct ?

(A) A dark spot will be formed at the point $\mathrm{P}_{2}$

(B) The angular separation between two consecutive bright spots decreases as we move from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$ along the first quadrant

(C) At $\mathrm{P}_{2}$ the order of the fringe will be maximum

(D) The total number of fringes produced between $P_{1}$ and $\mathrm{P}_{2}$ in the first quadrant is close to 3000

[JEE Advanced 2017]

Sol. (C,D)


Liquid Solution – JEE Main Previous Year Question of with Solutions

JEE Main Previous Year Question of Chemistry with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Chemistry will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

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Previous Years AIEEE/JEE Mains Questions

Q. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the folloowing statements is correct regarding the behaviour of the solution ?

(1) The solution is non-ideal, showing –ve deviation from Raoult’s law

(2) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law

(3) The solution formed is an ideal solution.

(4) The solutionis non-ideal, showing +ve deviation from Raoult’s law

[AIEEE-2009]

Sol. (4)

(A) n–heptone : Non Polar

(B) Ethanol : Polar

$\mathrm{F}_{\mathrm{A}-\mathrm{B}}<\mathrm{F}_{\mathrm{A}-\mathrm{A}}, \mathrm{F}_{\mathrm{B}-\mathrm{B}} \Rightarrow+$ deviation


Q. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively :-

(1) 400 and 600 (2) 500 and 600 (3) 200 and 300 (4) 300 and 400

[AIEEE-2009]

Sol. (1)

$550=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{4}+\mathrm{P}_{\mathrm{B}}^{\circ} \times \frac{3}{4}$

$560=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{5}+\mathrm{P}_{\mathrm{B}}^{\circ} \times \frac{4}{5}$

$\mathrm{P}_{\mathrm{A}}^{\circ}=400

\quad \mathrm{P}_{\mathrm{B}}^{\circ}=600$ torr


Q. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of

the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively.

Vapour pressure of the solution obtained by mixing 25.0 of heptane and 35 g of octane

will be (molar mass of heptane = 100 g $\mathrm{mol}^{-1}$ and of octane = 114 g $\mathrm{mol}^{-1}$) :-

(1) 144.5 kPa           (2) 72.0 kPa             (3) 36.1 kPa            (4) 96.2 kPa

[AIEEE-2010]

Sol. (2)

$\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{X}_{\mathrm{A}}=105 \times \frac{1 / 4}{1 / 4+0.307}$

$=105 \times 0.449=47.13 \mathrm{K} \mathrm{Pa}$

$\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{X}_{\mathrm{B}}=45 \times 0.551=24.795$

$\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}=71.925 \mathrm{atm}$


Q. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $\left(\Delta \mathrm{T}_{\mathrm{f}}\right)$, when 0.01 mol of sodium sulphate isdissolved in 1 kg of water, is $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right):$ :-

(1) 0.0186 K            (2) 0.0372 K              (3) 0.0558 K              (4) 0.0744 K

[AIEEE-2010]

Sol. (3)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m}$

$=3 \times 1.86 \times 0.01 / 1$

$=0.0558 \mathrm{K}$


Q. The molality of a urea solution in which 0.0100g of urea, $\left.\left[\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]$ is added to 0.3000 $\mathrm{dm}^{3}$ of water at STP is :-

(1) 0.555 m

(2) $5.55 \times 10^{-4} \mathrm{m}$

(3) 33.3 m

(4) $3.33 \times 10^{-2} \mathrm{m}$

[AIEEE-2011]

Sol. (2)

$\mathrm{m}=\frac{\mathrm{n}}{\mathrm{W}(\mathrm{kg})}=\frac{0.01 / 60}{0.3 \mathrm{kg}}=5.55 \times 10^{-4} \mathrm{mol} / \mathrm{kg}$


Q. A 5% solution of cane sugar (molar mass 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is :-

(1) 136.2           (2) 171.2           (3) 68.4            (4) 34.2

[AIEEE-2011]

Sol. (3)

$\pi_{\mathrm{c.s}}=\pi_{\mathrm{Unk}}$

$\left(\frac{\mathrm{n}}{\mathrm{V}}\right)_{\mathrm{c.s.}} \mathrm{RT}=\left(\frac{\mathrm{n}}{\mathrm{V}}\right)_{\mathrm{unk} .} \mathrm{RT}$

$\frac{5 \times 10}{342}=\frac{1 \times 10}{\mathrm{M}}$

M = 68.4 gm/mol


Q. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at – $6^{\circ} \mathrm{C}$ will be :

$\left(\mathrm{K}_{\mathrm{f}} \text { for water }=1.86 \mathrm{K} \mathrm{kgmol}^{-1}, \text { and molar mass of ethylene glycol }=62 \mathrm{gmol}^{-1}\right)$

(1) 400.00 g             (2) 304.60 g            (3) 804.32 g            (4) 204.30 g

[AIEEE-2011]

Sol. (3)

$6=1.86 \times \frac{\mathrm{w} / 62}{4} \Rightarrow \mathrm{w}=800 \mathrm{gm}$


Q. The degree of dissociation () of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :-

$(1) \alpha=\frac{\mathrm{x}+\mathrm{y}-1}{\mathrm{i}-1}$

(2) $\alpha=\frac{\mathrm{x}+\mathrm{y}+1}{\mathrm{i}-1}$

(3) $\alpha=\frac{\mathrm{i}-1}{(\mathrm{x}+\mathrm{y}-1)}$

(4) $\alpha=\frac{\mathrm{i}-1}{\mathrm{x}+\mathrm{y}+1}$

[AIEEE-2011]

Sol. (3)


Q. $\mathrm{K}_{\mathrm{f}}$ for water is 1.86 K kg $\mathrm{mol}^{-1}$. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to –$-2.8^{\circ} \mathrm{C} ?$

(1) 27 g            (2) 72 g            (3) 93 g             (4) 39 g

[AIEEE-2012]

Sol. (3)

$2.8=1.86 \times \frac{\mathrm{w} / 62}{1} \Rightarrow \mathrm{w}=93.33 \mathrm{gm}$


Q. A solution containing 0.85 g of $\mathrm{ZnCl}_{2}$ in 125.0 g of water freezes at $-0.23^{\circ} \mathrm{C}$ . The apparent degree of dissociation of the salt is :

($\mathbf{k}_{f}$ for water = 1.86 K kg $\mathrm{mol}^{-1}$, atomic mass ; Zn = 65.3 and Cl = 35.5)

(1) 1.36%            (2) 2.47%              (3) 73.5%             (4) 7.35%

[Jee (Main)-2012 online]

Sol. (3)

$0.23=(1+2 \alpha) \times 1.86 \times \frac{0.85 / 134.5}{0.125}$

$\alpha=0.735=73.5 \%$


Q. Liquids A and B form an ideal solution. At $30^{\circ}$C, the total vapour pressure of a solution containing 1 mol of A and 2 moles of B is 250 mm Hg. The total vapour pressure becomes 300 mm Hg when 1 more mol of A is added to the first solution. The vapour pressures of pure A and B at the same temperature are

(1) 450, 150 mm Hg

(2) 250, 300 mm Hg

(3) 125, 150 mm Hg

(4) 150, 450 mm Hg

[Jee (Main)-2012 online]

Sol. (1)

$250=\mathrm{P}_{\mathrm{A}}^{0} \times \frac{1}{3}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{2}{3}$

$300=\mathrm{P}_{\mathrm{A}}^{0} \times \frac{1}{2}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{1}{2}$

$\mathrm{P}_{\mathrm{A}}^{0}=450 \mathrm{mm}$

$\mathrm{P}_{\mathrm{B}}^{0}=150 \mathrm{mm}$


Q. The freezing point of a 1.00 m aqueous solution of HF is found to be $-1.91^{\circ} \mathrm{C}$. The

freezing point constant of water, $\mathrm{K}_{\mathrm{f}}$, is 1.86 K kg $\mathrm{mol}^{-1}$. The percentage dissociation of HF at this concentration is

(1) 2.7%             (2) 30%            (3) 10%             (4) 5.2%

[Jee (Main)-2012 online]

Sol. (1)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$1.91=(1+\alpha) \times 1.86 \times 1$

$\alpha=0.027$


Q. How many grams of methyl alcohol should be added to 10 litre tank of water to prevent its freezing at 268 K ?

$\left(\mathrm{K}_{f} \text { for water is } 1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$

(1) 899.04 g              (2) 886.02 g            (3) 868.06 g                 (4) 880.07 g

[Jee (Main)-2013 online]

Sol. (2)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}^{0}-\mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$273.15-268=1.86 \times \frac{\mathrm{w} / 32}{10}$

$\mathrm{w}=886.02 \mathrm{g}$


Q. Vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be :

(1) 0.137           (2) 0.205            (3) 0.237            (4) 0.435

[Jee (Main)-2013 online]

Sol. (3)

Benzen $\rightarrow 4$

Toluene $\rightarrow B$

y $_{B}=\frac{P_{B}^{0} \times X_{B}}{P_{B}^{0} X_{B}+P_{A}^{0} X_{A}}=\frac{37 \times 0.5}{37 \times 0.5+119 \times 0.5}=0.237$


Q. A molecule M associates in a given solvent according to the equation M  $(\mathrm{M})_{\mathrm{n}}$. For a certain concentration of M, the van’t Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is :

(1) 2              (2) 4               (3) 5               (4) 3

[Jee (Main)-2013 online]

Sol. (1)

$\mathrm{M}=\mathrm{M}_{\mathrm{n}}$

$1-0.2 \quad 0.2 / \mathrm{n}$

$0.9=\frac{1-0.2+0.2 / \mathrm{n}}{1}$

$0.9=0.8+\frac{0.2}{\mathrm{n}}$

$0.1=\frac{0.2}{\mathrm{n}}$

$\mathrm{n}=2$


Q. 12g of a nonvolatile solute dissolved in 108g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is :

(1) 60           (2) 80             (3) 40            (4) 20

[Jee (Main)-2013 online]

Sol. (4)

$\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}=0.1=\frac{12 / \mathrm{m}}{108 / 18} \Rightarrow \mathrm{m}=20$


Q. The molarity of a solution obtained by mixing 750 mL of 0.5(M)HCl with 250 mL of 2(M)HCl will be :-

(1) 0.875 M           (2) 1.00 M            (3) 1.75 M            (4) 0.975 M

[Jee (Main)-2013]

Sol. (1)

$\mathrm{M}_{\mathrm{f}}=\frac{\mathrm{M}_{1} \mathrm{V}_{1}+\mathrm{M}_{2} \mathrm{V}_{2}}{\mathrm{V}_{1}+\mathrm{V}_{2}}=0.875 \mathrm{M}$


Q. The observed osmotic pressure for a 0.10 M solution of Fe$\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2}$ at $25^{\circ} \mathrm{C}$ is 10.8 atm. The expected and experimental (observed) values of Van’t Hoff factor (i) will be respectively : $\left(\mathrm{R}=0.082 \mathrm{L} \mathrm{atm} \mathrm{k}^{-} \mathrm{mol}^{-1}\right)$

(1) 3 and 5.42          (2) 5 and 3.42           (3) 4 and 4.00           (4) 5 and 4.42

[Jee (Main)-2014 online]

Sol. (4)

$\pi_{\mathrm{ob}}=\mathrm{i} \frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}$

$10.8=\mathrm{i} \times 0.1 \times 0.082 \times 298$

$\mathrm{i}=4.42$


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) $\mathrm{A}-\mathrm{A}, \mathrm{B}-\mathrm{B}$ and $\mathrm{A}-\mathrm{B}$ interactions are identical

(3) $\mathrm{A}-\mathrm{B}$ interaction is stronger than $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$ interactions

(4) $\Delta \mathrm{H}_{\text {mixing }}>0$ (zero)

[Jee(Main)-2014 online]

Sol. (2)

$\Delta \mathrm{H}_{\operatorname{mix}}=0$


Q. Consider separate solution of $0.500 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}), 0.100 \mathrm{MM} \mathrm{g}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{aq}), 0.250 \mathrm{M} \mathrm{KBr}(\mathrm{aq})$ and 0.125 M $\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})$ at $25^{\circ} \mathrm{C}$. Which statement is true about these solutions, assuming all salts to be strong electrolytes ?

(1) 0.125 M $\mathrm{Na}_{3} \mathrm{PO}_{4}$ (aq) has the highest osmotic pressure.

(2) 0.500 M $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ (aq) has the highest osmotic pressure.

(3) They all have the same osmotic pressure.

(4) 0.100 M $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ (aq) has the highest osmotic pressure.

[Jee (Main)-2014]

Sol. (3)


Q. Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by :

(1) association

(2) dissociation

(3) complex formation

(4) partial ionization

[Jee (Main)-2015 online]

Sol. (1)

Acetic acid in non polar solvent (benzene) associates.


Q. A solution at $20^{\circ} \mathrm{C}$ is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively :

(1) 38.0 torr and 0.589

(2) 30.5 torr and 0.389

(3) 35.8 torr and 0.280

(4) 35.0 torr and 0.480

[Jee (Main)-2015 online]

Sol. (1)

$\begin{aligned} \mathrm{P}_{\mathrm{T}} &=\mathrm{P}_{\mathrm{A}}^{0} \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0} \mathrm{X}_{\mathrm{B}} \\ &=747 \times \frac{1.5}{5}+22.3 \times \frac{3.5}{5} \\ &=38 \mathrm{torr} \end{aligned}$


Q. The vapour pressure of acetone at $20^{\circ}$C is 185 torr. When 1.2 g of non-volatile substance was dissolved in 100 g of acetone at $20^{\circ}$$20^{\circ}$C, its vapour pressure was 183 torr. The molar mass $\left(\mathrm{g} \mathrm{mol}^{-1}\right)$ of the substance is :

(1) 128 (2) 488 (3) 32 (4) 64

[Jee (Main)-2015]

Sol. (4)

$\begin{array}{rl}{\frac{185-183}{185}} & {=\frac{1.2 / \mathrm{m}}{100 / 58}} \\ {\mathrm{m}=64} & {\mathrm{gm} / \mathrm{mol}}\end{array}$


Q. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ?

(1) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$

(2) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} .2 \mathrm{H}_{2} \mathrm{O}$

(3) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$

(4) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$

[Jee (Main)-2018]

Sol. (3)


Mind Map of Sound Waves Class 11, JEE & NEET – Download from here

Sound Waves in Class 11 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

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Mind Maps for Magnetic Effects of Current Revision – Class XII, JEE, NEET

Magnetic Effects od Current in Class 12th comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulae and important key concepts on finger tips.

 

 

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Magnetic Effect of current Mind Map

 

 


 

Mind Maps for Kinematics 1D Revision – Class XI, JEE, NEET

Kinematics 1D Chapter in Class 11 comprises variety of cases with important formulae and key points. So here are the mind maps to help you in remembering all the formulas and important key concepts on finger tips.

 

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Caboxylic Acid – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. Match each of the compound in Column I with its characteristic reaction(s) in Column II.

[IIT 2009]

Sol.


Q. Match each of the compounds given in Column I with the reaction(s), that they can undergo, given in Column II.

[IIT 2009]

Sol. ($(A) \rightarrow P, Q, S, T:(B) \rightarrow P, S, T ;(C) \rightarrow P:(D) \rightarrow R$)


Q. The major product of the following reaction is

[IIT 2011]

Sol. ($(\mathrm{A}) \rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{T} ;(\mathrm{B}) \rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{S}, \mathrm{T},(\mathrm{C}) \rightarrow \mathrm{R}, \mathrm{S}, ;(\mathrm{D}) \rightarrow \mathrm{P}$)


Q. With reference the scheme given, which of the given statement(s) about T, U, V & W is (are) correct ?

(A) ‘T’ is soluble in hot aq NaOH

(B) ‘U’ is optically active

(C) mol formula of $\mathrm{W}$ is $\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}_{4}$

(D) V gives effervescence with aq NaHCO $_{3}$

[IIT 2012]

Sol. (A)


Q. Identify the binary mixtures (s) that can be separated into the individual compounds, by differential extraction, as shown in the given scheme –

[IIT 2012]

Sol. (A,C,D)


Q. The total number of carboxylic acid groups in the product P is

[IIT 2013]

Sol. (B,D)


Q. In the reaction shown below, the major product(s) formed is / are :

[IIT 2014]

Sol. 2


Q. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List-I with an appropriate structure from List-II and select the correct answer using the code given below the lists.

[IIT 2014]

Sol. (A)


Q. The major product of the reaction is :

[IIT 2015]

Sol. (A)


Q. Aniline reacts with mixed acid (conc. HNO, and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ ) at 288 K to give P (51%), Q (47%) and R (2%). The major product(s) the following reaction sequence is (are) :-

[JEE Adv. 2018]

Sol. (C)


Q. In the following reaction sequence, the correct structure(s) of X is (are)

[JEE Adv. 2018]

Sol. (D)


Treatment of benzene with CO/HCl in the presence of anhydrous $\mathrm{AlCl}_{3} / \mathrm{CuCl}$ followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with

$\mathrm{Br}_{2} / \mathrm{Na}_{2} \mathrm{CO}_{3}$, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with $\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{C}$, followed by $\mathrm{H}_{3} \mathrm{PO}_{4}$ treatment gives Z as the major product.

(There are two questions based on PARAGRAPH “X”, the question given below is one of them)

Q. The compound Y is :-

[JEE Adv. 2018]

Sol. (B)


Q. The compound Z is :-

[JEE Adv. 2018]

Sol. (C)


An organic acid P $\left(\mathrm{C}_{1} \mathrm{H}_{12} \mathrm{O}_{2}\right)$ can easily be oxidized to a dibasic acid which reacts with ethyleneglycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S.

(There are two questions based on PARAGRAPH “A”, the question given below is one of them)

Q.

[JEE Adv. 2018]

Sol. (A)


Q. The compound R is

(A)

(B)

(C)

(D)

[JEE Adv. 2018]

Sol. (A)


Q. The compound S is

[JEE Adv. 2018]

Sol. (B)


 

Radioactivity – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. The total number of $\alpha$ and $\beta$particles emitted in the nuclear reaction $_{92}^{238} \mathrm{U} \rightarrow_{82}^{214} \mathrm{Pb}$ is

[JEE 2009]

Sol. 8


Q. The number of neutrons emitted when $_{92}^{235} \mathrm{U}$ undergoes controlled nuclear fission to $_{54}^{142} \mathrm{Xe}$ and is –

[JEE 2010]

Sol. 4


Q. Bombardment of aluminium by $\alpha$ -particle leads to its artificial disintegration in two ways,

(i) and (ii) as shown. Products X, Y and Z respectively are :

(A) proton, neutron, positron

(B) neutron, positron, proton

(C) proton, positron, neutron

(D) positron, proton, neutron

[JEE 2011]

Sol. (A)


Q. The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group , element X belongs in the periodic table ?

[JEE 2012]

Sol. 8


Q. In the nuclear transmutation

(X, Y) is(are)

$(\mathrm{A})(\gamma, \mathrm{n})$

(B) (p, D)

(C) (n, D)

$(\mathrm{D})(\gamma, \mathrm{D})$

[JEE 2013]

Sol. (AB)


Q. A closed vessel with rigid walls contains 1 mol of 238

92 $\mathrm{U}$ U and 1 mol of air at 298 K. Considering complete decay of $^{238}_{92} \mathrm{U}$ the ratio of the final pressure to the initial pressure of the system at 298 K is –

[JEE 2015]

Sol. 9


Metallurgy – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Previous Years JEE Advance Questions

Paragraph for questions 1 to 3

Copper is the most nobel of the first row transition metals and occurs in small deposits in several

countries. Ores of copper include chalcanthite $\left(\mathrm{CuSO}_{4}, 5 \mathrm{H}_{2} \mathrm{O}\right),$ atacamite $\left(\mathrm{Cu}_{2} \mathrm{Cl}(\mathrm{OH})_{3}\right),$ cuprite $\left(\mathrm{Cu}_{2} \mathrm{O}\right),$ copper glance (Cu.S) and malachite $\left(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{CO}_{3}\right) .$ However, $80 \%$ of the world copper production comes from the ore chalcopyrite (CuFeS_{2} ) . \text { The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.

Q. Partial roasting of chalcopyrite produces :-

(A) $\mathrm{Cu}_{2} \mathrm{S}$ and $\mathrm{FeO}$

(B) $\mathrm{Cu}_{2} \mathrm{O}$ and FeO

(C) $\mathrm{CuS}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(D) $\mathrm{Cu}_{2} \mathrm{O}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3}$

[JEE-2010]

Sol. (A)


Q. Iron is removed from chalcopyrite as :-

(A) FeO

(B) FeS

(C) $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(D) FeSiO $_{3}$

[JEE-2010]

Sol. (D)


Q. In self-reduction, the reducing species is :-

(A) S

(B) $\mathrm{O}^{2-}$

(C) $\mathrm{S}^{2-}$

(D) $\mathrm{SO}_{2}$

[JEE-2010]

Sol. (C)


Q. Extraction of metal from the ore cassiterite involves –

(A) carbon reduction of an oxide ore

(B) self-reduction of a sulphide ore

(C) removal of copper impurity

(D) removal of iron impurity

[JEE-2011]

Sol. (A,C,D)

$\mathrm{SnO}_{2}+2 \mathrm{C} \rightarrow \mathrm{Sn}+2 \mathrm{CO}$

Cassetrite contains impurity of Wolframite (FeWO,), Which is a ferro magnatic.


Q. Oxidation states of the metal in the minerals haematite and magnetite, respectively, are –

(A) II, III in haematite and III in magnetite

(B) II, III in haematite and II in magnetite

(C) II in haematite and II, III in magnetite

(D) III in haematite and II, III in magnetite

[JEE-2011]

Sol. (D)

Haematite Ore $\Rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3} ;$ Oxidation state $(+3)$ magnetite ore $\Rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}\left(\mathrm{FeO}+\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=$ Oxidation

state of $(+2 \&+3)$


Q. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are :

(A) $\mathrm{O}_{2}$ and CO respectively.

(B) $\mathrm{O}_{2}$ and $\mathrm{Zn}$ dust respectively.

(C) $\mathrm{HNO}_{3}$ and $\mathrm{Zn}$ dust respectively.

(D) $\mathrm{HNO}_{3}$ and CO respectively.

[JEE-2012]

Sol. (B)

$2 \mathrm{Ag}+4 \mathrm{NaCN}+\mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}(\text { air }) \longrightarrow 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]+2 \mathrm{NaOH},$ Here $\mathrm{O}_{2}$ is oxidising agent $\& \mathrm{Zn}^{-}$

dust act as reducing agent.


Q. Sulfide ores are common for the metals –

(A) Ag, Cu and Pb

(B) Ag, Cu and Sn

(B) Ag, Cu and Sn

(D) Al, Cu and Pb

[JEE-2013]

Sol. (A)


Q. The carbon-based reduction method is NOT used for the extraction of

(A) $\operatorname{tin}$ from $\mathrm{SnO}_{2}$

(B) Iron from $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(C) aluminium from $\mathrm{Al}_{2} \mathrm{O}_{3}$

(D) magnesium from $\mathrm{MgCO}_{3} . \mathrm{CaCO}_{3}$

[JEE-2013]

Sol. (C,D)

Being a more reactive metal “Al” and “Mg” can be reduced by electrolytic reduction.


Q. Upon heating with $\mathrm{Cu}_{2} \mathrm{S},$ the reagent(s) that give copper metal is/are

(A) CuFeS $_{2}$

(B) CuO

(C) $\mathrm{Cu}_{2} \mathrm{O}$

(D) $\mathrm{CuSO}_{4}$

[JEE Adv. 2015]

Sol. (A,C,D)


Q. Copper is purified by electrolytic refining of blister copper. The correct statement(s) about this process is (are) –

(A) Impure Cu strip is used as cathode

(B) Acidified aqueuous $\mathrm{CuSO}_{4}$ is used as electrolyte

(C) Pure Cu deposits at cathode

(D) Impurities settle as anode-mud

[JEE Adv. 2015]

Sol. (B,C,D)

Impure “Cu” act as a Anode . While ” Pure thin strips of Cu act as cathode and aqueous solution of $\mathrm{CuSO}_{4}$ act as electrolyte.


Q. Match the anionic species given in Column-I that are present in the ore(s) given in

Column-II –

[JEE Adv. 2015]

Sol. $(A \rightarrow P, Q, S, B \rightarrow T, C \rightarrow Q, R, D \rightarrow R)$

Name of ores & their metals.


Q. Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnance such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of $\mathrm{O}_{2}$ consumed is ______ .

(Atomic weights in $\left.\mathrm{g} \text { mol }^{-1}: \mathrm{O}=16, \mathrm{S}=32, \mathrm{Pb}=207\right)$

[JEE Adv. 2018]

Sol. 6.47


Liquid Solution – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. The Henry’s law constant for the solubility of $\mathrm{N}_{2}$ gas in water at 298 K is 1.0 × $10^{5}$ atm. The mole fraction of N2 in air is 0.8. The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of waterat 298 K and 5 atm pressure is-

(A) $4.0 \times 10^{-4}$

(B) $4.0 \times 10^{-5}$

(C) $5.0 \times 10^{-4}$

(D) $4.0 \times 10^{-5}$

[JEE 2009]

Sol. (A)

$\mathrm{P}_{\mathrm{N}_{2}}=\mathrm{K}_{\mathrm{H}} \mathrm{X}_{\mathrm{N}_{2}}$

$\mathrm{Y}_{\mathrm{N}_{2}} \cdot \mathrm{P}_{\mathrm{T}}=\mathrm{K}_{\mathrm{H}} \times \mathrm{N}_{2}$

$0.8 \times 5=1 \times 10^{5} \times \frac{\mathrm{n}}{\mathrm{n}+10}$

$4=10^{5} \times \frac{\mathrm{n}}{10}$

$\mathrm{n}=4 \times 10^{-4}$


Q. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2^{\circ} \mathrm{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is-(take $\left.\mathrm{K}_{\mathrm{b}}=0.76 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$

(A) 724             (B) 740            (C) 736             (D) 718

[JEE 2011]

Sol. (A)


Q. The freezing point (in °C) of a solution containing 0.1 g of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ (Mol. Wt. 329) in

100 g of water $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$ is –

(A) $-2.3 \times 10^{-2}$

(B) $-5.7 \times 10^{-2}$

(C) $-5.7 \times 10^{-3}$

(D)-1.2 \times 10^{-2}

[JEE 2011]

Sol. (A)

$\mathrm{T}_{\mathrm{f}}^{\prime}=\mathrm{T}_{\mathrm{f}}-\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}-\mathrm{i} \mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$

$=0^{\circ} \mathrm{C}-4 \times 1.86 \times \frac{0.1 / 329}{100 / 1000}$

$=-0.023^{\circ} \mathrm{C}=-2.3 \times 10^{2} \mathrm{C}$


Q. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2^{\circ} \mathrm{C}. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is \text { (take }\left.\mathrm{K}_{\mathrm{b}}=0.76 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)

(A) 724 (B) 740 (C) 736 (D) 718

[JEE 2012]

Sol. (A)

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m}$

$2=0.76 \times \frac{\mathrm{n}}{100 / 1000}$

$\mathrm{n}=0.263 \mathrm{mol}$

$\mathrm{P}_{\mathrm{S}}=\mathrm{P}^{\circ} \mathrm{x}_{\text {solvent }}=760 \times \frac{\frac{100}{18}}{\frac{100}{18}+0.263}=760 \times \frac{5.55}{5.82}=724.7$ torr


Q. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is(are)

(A) $\Delta \mathrm{G}$ is positive

(B) $\Delta S_{\text {system }}$ is positive

(C) $\Delta \mathrm{S}_{\text {surroundings }}=0$

(D) $\Delta \mathrm{H}=0$

[J-Adv. 2013]

Sol. (B,C,D)


Q. A compound $\mathrm{H}_{2} \mathrm{X}$ with molar weight of 80 g is dissolved in a solvent having density of

$0.4 \mathrm{g} \mathrm{mL}^{-1}$, Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

[JEE-Adv. 2014]

Sol. 8

$\mathrm{m}=\frac{3.2 \mathrm{mol}}{0.4 \mathrm{Kg}}=8 \mathrm{mol} / \mathrm{kg}$


Q. If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong eletrolyte) is $-0.0558^{\circ} \mathrm{C}$ , the number of chloride (s) in the coordination sphere of the complex is- $\left[\mathrm{K}_{\mathrm{f}} \text { of water }=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right]$

[JEE-Adv. 2015]

Sol. 1

0.0558 = i × 1.86 × 0.01

i = 3

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$


Q. Mixture(s) showing positive deviation from Raoult’s law at $35^{\circ}$C is (are)

(A) carbon tetrachloride + methanol

(B) carbon disulphide + acetone

(C) benzene + toluene

(D) phenol + aniline

[JEE – Adv. 2016]

Sol. (A,C)

(A) H-bonding of methanol breaks when $\mathrm{CCl}_{4}$ is added so bonds become weaker, resulting positive

deviation.

(B) Mixing of polar and non-polar liquids will produce a solution of weaker interaction, resulting positive deviation

(C) Ideal solution

(D) –ve deviation because stronger H-bond is formed.


Q. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure, Here $\mathrm{x}_{\mathrm{L}}$ and $\mathbf{X}_{\mathrm{M}}$ represent mole fractions of L and M, respectively, in the solution. the correct statement(s) applicable to this system is(are) –

(A) Attractive intramolecular interactions between L–L in pure liquid L and M–M in pure liquid M are stronger than those between L–M when mixed in solution

(B) The point $Z$ represents vapour pressure of pure liquid $\mathrm{M}$ and Raoult’s law is obeyed when $\quad \mathrm{x}_{\mathrm{L}} \rightarrow 0$

(C) The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when $\mathrm{x}_{\mathrm{L}} \rightarrow 1$

(D) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed from $\mathrm{x}_{\mathrm{L}}=0$ to $\mathrm{x}_{\mathrm{L}}=1$

[JEE – Adv. 2017]

Sol. (A,C)


Q. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg $\mathrm{mol}^{-1}$ . The figures shown below represents plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is 46 g $\mathrm{mol}^{-1}$]

Among the following, the option representing change in the freezing point is –

[JEE – Adv. 2017]

Sol. (D)

Ethanol should be considered non volatile as per given option

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$\Delta \mathrm{T}_{\mathrm{f}}=2 \times \frac{34.5}{46 \times 0.5}=3 \mathrm{K}$


Q. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions $\mathbf{X}_{\mathrm{A}}$ and $\mathrm{x}_{\mathrm{B}}$, respectively, has vapour pressure of 22.5 Torr. The value of $\mathrm{x}_{\mathrm{A}} / \mathrm{x}_{\mathrm{B}}$ in the new solution is____.

(given that the vapour pressure of pure liquid A is 20 Torr at temperature T)

[JEE – Adv. 2018]

Sol. 19

$45=\mathrm{P}_{\mathrm{A}}^{\mathrm{o}} \times \frac{1}{2}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \times \frac{1}{2}$$\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}+\mathrm{P}_{\mathrm{B}}^{\circ}=90 \ldots \ldots(1)$

given $\mathrm{P}_{\mathrm{A}}^{\circ}=20$ torr

$\mathrm{P}_{\mathrm{B}}^{\circ}=70 \mathrm{torr}$

$\Rightarrow 22.5$ torr $=20 \mathrm{x}_{\mathrm{A}}+70\left(1-\mathrm{x}_{\mathrm{A}}\right)$

$=70-50 \mathrm{x}_{\mathrm{A}}$

$\mathrm{x}_{\mathrm{A}}=\left(\frac{70-22.5}{50}\right)=0.95$

$\mathrm{x}_{\mathrm{B}}=0.05$

So $\frac{\mathrm{x}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{B}}}=\frac{0.95}{0.05}=19$


Q. The plot given below shows P–T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.

On addition of equal number of moles a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is ___.

[JEE – Adv. 2018]

Sol. 0.05

From graph


Isomerism – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. The correct statement(s) about the compound $\mathrm{H}_{3} \mathrm{C}$ (HO)HC – CH = CH – CH(OH) $\mathrm{CH}_{3}$ (X) is (are) :

(A) The total number of stereoisomers possible for X is 6

(B) The total number of diastereomers possible for X is 3

(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4

(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2

[iit-2009]

Sol. (A,D)


Q. In the Newman projection for 2,2–dimethylbutane –

X and Y can respectively be –

(A) H and H

(B) $\mathrm{H}$ and $\mathrm{C}_{2} \mathrm{H}_{5}$

(C) $\mathrm{C}_{2} \mathrm{H}_{5}$ and $\mathrm{H}$

(D) $\mathrm{CH}_{3}$ and $\mathrm{CH}_{3}$

[iit-2010]

Sol. (B,D)


Q. Amongst the given option, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are) –

[iit-2011]

Sol. (B,C)


Q. Which of the given statement(s) about N,O,P and Q with respect to M is (are) correct ?

(A) M and N are non-mirror image stereoisomers

(B) M and O are identical

(C) M and P are enantiomers

(D) M and Q are identical

[JEE-2012]

Sol. (A,B,C)


Q. The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are) :

[JEE-2014]

Sol. 3


ElectroChemistry – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. For the reaction of $\mathrm{NO}_{3}^{-}$ ion in an aqueous solution, $\mathrm{E}^{\circ}$ is +0.96 V. Values of $\mathrm{E}^{\circ}$ for some metal ions are given below

The pair(s) of metal that is(are) oxidised by $\mathrm{NO}_{3}^{-}$ in aqueous solution is(are)

(A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V

[JEE 2009]

Sol. (A,B,D)

(A,B,D) as $\mathrm{E}^{\circ}$ will be positive


Paragraph for Questions 2 to 3

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :

$\mathrm{M}(\mathrm{s}) | \mathrm{M}^{+}\left(\mathrm{aq} ; 0.05 \text { molar) } \| \mathrm{M}^{+}(\mathrm{aq} ; 1 \mathrm{molar}) | \mathrm{M}(\mathrm{s})\right.$

For the above electrolytic cell the magnitude of the cell potential $\left|\mathrm{E}_{\mathrm{cell}}\right|$ = 70 mV.

Q. For the above cell :-

(A) $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}>0$

(B) $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}<0$

(C) $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}^{0}>0$

(D) $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}^{\text {o }}<0$

[JEE 2010]

Sol. (C)

$\mathrm{E}_{1}=-\frac{059}{1} \log \frac{05}{1}=(+) \mathrm{ve} \Rightarrow \mathrm{so}$


Q. If the 0.05 molar solution of $\mathrm{M}^{+}$ is replaced by a 0.0025 molar $\mathrm{M}^{+}$ solution, then the magnitude of the cell potential would be :-

(A) 35 mV              (B) 70 mV             (C) 140 mV              (D) 700 mV

[JEE 2010]

Sol. (B)


Q. Consider the following cell reaction :

$2 \mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}(\ell) \quad \mathrm{E}^{\circ}=1.67 \mathrm{V}$

$\mathrm{At}\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_{2}\right)=0.1$ atm and $\mathrm{pH}=3,$ the cell potential at $25^{\circ} \mathrm{C}$ is –

(A) 1.47 V (B) 1.77 V (C) 1.87 V (D) 1.57 V

[JEE 2011]

Sol. (D)


Q. $\mathrm{AgNO}_{3}$ (aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. the plot of conductance () versus the volume of $\mathrm{AgNO}_{3}$ is –

(A) (P) (B) (Q) (C) (R) (D) (S)

[JEE 2011]

Sol. (D)


Paragraph for Question 6 and 7

The electrochemical cell shown below is a concentration cell.

$\mathrm{M} | \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt,

$\mathrm{MX}_{2}$) | | $\mathbf{M}^{2+}$ (0.001 mol $\mathrm{dm}^{-3}$) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059V.

Q. The value of G $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ for the given cell is (take If = 96500 C $\mathrm{mol}^{-1}$)

(A) –5.7

(B) 5.7

(C) 11.4

(D) –11.4.

[JEE 2012]

Sol. (B)


Q. The solubility product $\left(\mathrm{K}_{\mathrm{sp}} ; \mathrm{mol}^{3} \mathrm{dm}^{-9}\right)$ of $\mathrm{MX}_{2}$ at 298 K based on the information available for the given concentration cell is (take 2.303 × R × 298/F = 0.059 V)

(A) $1 \times 10^{-15}$

(B) $4 \times 10^{-15}$

(C) $1 \times 10^{-12}$

(D) $1 \times 10^{-12}$

Sol. (D)

$\Delta \mathrm{G}=\frac{-2 \times 96500 \times .059}{1000}=11.4$


Q. The standard reduction potential data at $25^{\circ} \mathrm{C}$ is given below

Match $\mathrm{E}^{\mathrm{o}}$ of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below the lists :

[JEE-Adv. 2013]

Sol. (D)


Q. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List-I. The variation in conductivity of these reactions is given in List-II. Match List-I with List-II and select the correct answer using the code given below the lists :

[JEE-Adv. 2013]

Sol. (A)

$\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{N}+\mathrm{CH}_{3} \mathrm{COOH} \Rightarrow$ Weak acid and weak base so conductivity increases and then does not change much so option 3 hence and (a)


Q. In a galvanic cell , the salt bridge –

(A) Does not participate chemically in the cell reaction

(B) Stops the diffusion of ions from one electrode to another

(C) Is necessary for the occurence of the cell reaction

(D) Ensures mixing of the two electrolytic solutions

[JEE-Adv. 2014]

Sol. (A,B)

Fact


Q. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.1 M). If $\lambda_{\mathrm{X}^{-}}^{0} \approx \lambda_{\mathrm{Y}^{-}}^{0}$ the difference in their $\mathrm{pK}_{\mathrm{a}}$ values $, \mathrm{pK}_{\mathrm{a}}(\mathrm{HX})-\mathrm{p} \mathrm{K}_{\mathrm{a}}(\mathrm{HY}),$ is (consider degree of ionization of both acids to be <<1).

[JEE-Adv. 2015]

Sol. 3


Q. All the energy released from the reaction X $\rightarrow \mathrm{Y}, \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-193 \mathrm{kJ} \mathrm{mol}^{-1}$ is used for the oxidizing $\mathrm{M}^{+}$ and $\mathrm{M}^{+} \rightarrow \mathrm{M}^{3+}+2 \mathrm{e}^{-}, \mathrm{E}^{\circ}=-0.25 \mathrm{V}$

Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is $\left[\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right]$

[JEE-Adv. 2015]

Sol. 4


Q. For the electrochemical cell,

$\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})$

the standard emf of the cell is 2.70 V at 300 K. When the concentration of $\mathrm{Mg}^{2+}$ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is____.

(given, $\frac{\mathrm{F}}{\mathrm{R}}=11500 \mathrm{KV}^{-1}$ where F is the Faraday constant and R is the gas constant, ln(10) = 2.30)

[JEE-Adv. 2018]

Sol. 10


Q. Consider an electrochemical cell: $\mathrm{A}(\mathrm{s})\left|\mathrm{A}^{\mathrm{n}+}(\mathrm{aq}, 2 \mathrm{M}) \| \mathrm{B}^{2 \mathrm{n}+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{B}(\mathrm{s})$. The value of $\Delta \mathrm{H}^{\theta}$ for the cell reaction is twice that of $\Delta \mathrm{G}^{\theta}$ at 300 K. If the emf of the cell is zero, the $\Delta \mathrm{S}^{\theta}$ $\left(\text { in } \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)$ of the cell reaction per mole of B formed at 300 K is___.
(Given : ln (2) = 0.7, R (universal gas constant) = $8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. H, S and G are enthalpy, entropy and Gibbs energy, respectively.)

[JEE-Adv. 2018]

Sol. (-11.62)


Chemical Kinetics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. Plots showing the variation of the rate constant (k) with temperature (T) are given

below. The plot that follows Arrhenius equation is –

[JEE 2010]

Sol. (A)

Arrhenius equation : $\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

(taking A and $\mathrm{E}_{\mathrm{a}}$ to be constant, differentiating w.r.t. T)

$\frac{\mathrm{d} \mathrm{k}}{\mathrm{d} \mathrm{T}}=\left[\frac{\mathrm{AE}_{\mathrm{a}}}{\mathrm{RT}^{2}}\right] \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

assuming A and $\mathrm{E}_{\mathrm{a}}$ to be constant theoretically, the plot should be

ButNo such option is given. Now experimentally, A and $\mathrm{E}_{\mathrm{a}}$ both vary with temperature and k increases as T increases and become very large at infinite temperature. Hence option (A) is correct.


Q. The concentration of R in the reaction $\mathrm{R} \rightarrow \mathrm{P}$ was measured as a function of time and the following data is obtained :

[JEE 2010]

Sol. 0

Zero order reaction because rate is constant with time.

Alternative solution : By hit and trial method, assuming the reaction is of zero order, putting given data in integrated expression for zero order.

the value of k is same from different data so reaction is zero order reaction.


Q. For the first order reaction

$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$

(A) the concentration of the reactant decreases exponentially with time

(B) the half-life of the reaction decreases with increasing temperature.

(C) the half-life of the reaction depends on the initial concentration of the reactant.

(D) the reaction proceeds to 99.6% completion in eight half-life duration.

[JEE 2011]

Sol. (A,B,D)

$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_{0} \mathrm{e}^{-\mathrm{kt}}$

for option (D)

$\frac{1}{\mathrm{t}_{1 / 2}} \ln \frac{100}{50}=\frac{1}{\mathrm{t}_{99.6 \%}} \ln \frac{100}{0.4}$


Q. An organic compound undergoes first-order decomposition . The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are $$

\mathrm{t}_{1 / 8} \text { and } \mathrm{t}_{1 / 10}

$$ What is the value of $$

\frac{\left[\mathrm{t}_{1 / 8}\right]}{\mathrm{t}_{1 / 10}} \times 10 ?\left(\text { take } \log _{10} 2=0.3\right)

$$

[JEE 2012]

Sol. 9


Q. In the reaction :

$$

\mathrm{P}+\mathrm{Q} \longrightarrow \mathrm{R}+\mathrm{S}

$$

the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The

concentration of Q varies with reaction time as shown in the figure. The overall order of the

reaction is –

(A) 2 (B) 3 (C) 0 (D) 1

[JEE 2013]

Sol. 9


Q. For the elementary reaction $$

\mathrm{M} \rightarrow \mathrm{N}$$, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

(A) 4 (B) 3(C) 2(D) 1

[JEE 2014]

Sol. (D)


Q. In dilute aqueous $$

\mathrm{H}_{2} \mathrm{SO}_{4}$$ , the complex diaquodioxalatoferrate(II) is oxidized by $\mathrm{MnO}_{4}^{-}$.For this reaction, the ratio of the rate of change of $$

\left[\mathrm{H}^{+}\right]$$ to the rate of change of [MnO4–] is.

[JEE 2015]

Sol. (B)


Q. The % yield of ammonia as a function of time in the reaction

If this reaction is conducted at $\left(P, T_{2}\right)$, with $\mathrm{T}_{2}>\mathrm{T}_{1}$, the % yield of ammonia as a function of time is represented by –

[JEE 2015]

Sol. (B)

$\therefore$ % yield will increase in initial stages due to increase in net speed

As time proceeds $\Rightarrow \mathrm{r}_{\mathrm{net}}=\mathrm{k}_{\mathrm{f}}\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}-\mathrm{k}_{\mathrm{b}}\left[\mathrm{NH}_{3}\right]^{2}$

On increasing temp., $\mathrm{k}_{\mathrm{f}} \& \mathrm{k}_{\mathrm{b}}$ increase

but increase of $\mathrm{k}_{\mathrm{b}}$ is more

so % yield will decrease

% yield will increase in initial stage due to enhance speed but as time proceeds , final yield is governed by thermodynamics due to which yield decrease since reaction is exothermic


Q. For a first order reaction A(g)  2B(g) + C(g) at constant volume and 300 K, the

total pressure at the beginning (t = 0) and at time t are P0 and Pt, respectively. Initially, only A

is present with concentration [A]0, and $t_{1 / 3}$ is the time required for the partial pressure

of A to reach $1 / 3^{\mathrm{rd}}$ of its initial value. The correct option(s) is (are) :-

(Assume that all these gases behave as ideal gases)

[JEE Advance 2018]

Sol. (A,D)


Q. Consider the following reversible reaction,

$\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \square \mathrm{AB}(\mathrm{g})$

The activition energy of the backward reaction exceeds that of the forward reaction by 2RT $\left(\text { in } \mathrm{J} \mathrm{mol}^{-1}\right)$. If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta \mathrm{G}^{\theta}$(in J mol–1) for the reaction at 300 K is____.

(Given ; ln (2) = 0.7, RT = 2500 J $\mathrm{mol}^{-1}$ at 300 K and G is the Gibbs energy)

[JEE Advance 2018]

Sol. 8500