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) • September 11, 2021 at 1:17 pm

Nice prb sir … 👍

0
• September 3, 2021 at 12:02 pm

Everyone should try again this type of questions

0
• August 31, 2021 at 8:16 pm

RK BRO IS ROCKING YOU KNOW

1
• August 31, 2021 at 8:16 pm

RK BRO IS ROCKING

1
• August 12, 2021 at 12:27 pm

Thanks u really made my day

0
• August 12, 2021 at 12:20 pm

really helpfull to make a strong grip in chemistry for jee mains

0
• June 12, 2021 at 7:08 am

thanks boss

4
• June 12, 2021 at 7:21 am

I am unable to.do maximum question what should i do ???

101
• May 22, 2021 at 2:40 am

6no sawal ka solution koi batao plz☹️

17
• June 17, 2021 at 7:44 pm

i am also looking for same now u got the ans?

0
• April 29, 2021 at 10:47 am

thank you

6
• April 17, 2021 at 4:02 pm

maja aa gya

5
• February 2, 2021 at 11:55 am

acha hai re

4
• February 2, 2021 at 11:54 am

bc

3
• December 10, 2020 at 5:12 pm

very ueful keep some more and more questions..thanku

29
• October 26, 2020 at 12:36 pm

wtf

6
• November 24, 2020 at 7:01 pm

Wt happen dude

10
• October 21, 2020 at 10:33 am

When I click on solution it said that there is error.please solve that problem

3
• October 15, 2020 at 3:38 pm

Nice
Awesome work👍

12
• October 9, 2020 at 11:58 pm

It’s good 👍

8
• October 5, 2020 at 4:30 pm

0
• February 24, 2021 at 7:53 am

Dil khus krdiya ,, After 4 hr , i have to attempt JEE MAINS😂

5
• September 14, 2020 at 10:07 am

Thanks to give this type of questions

0
• September 2, 2020 at 11:37 am

sir pc me aapka aap kaise install kare

0
• January 17, 2021 at 3:39 pm

Pehley spelling likhna sikh lo

110
• August 17, 2020 at 1:28 pm

thanks and good

1
• August 17, 2020 at 1:28 pm

thanks and good
after this information i have a AIR 621 in jee mains.

2
• July 10, 2021 at 10:21 pm

Hey plz give some advice how to study for jee as i am in 11th

0
• August 10, 2020 at 6:54 pm

1
• July 26, 2020 at 11:28 am

very useful

0
• July 3, 2020 at 8:01 pm

So good intiative hai sir

1
• June 24, 2020 at 4:10 pm

Thanks a lot

5
• June 2, 2020 at 9:20 pm

good

0
• May 25, 2020 at 4:13 pm

please provide me daily tough questions {DPP} of IIT JEE {mains} and {advance} on my email ID ; [email protected]

0
• May 18, 2020 at 4:25 pm

It’s a helpful site to JEE and NEET aspirants 🙏

0
• May 16, 2020 at 4:00 pm

How many questions are asked in JEE mains 2020 from mole concept ??

2
• June 24, 2020 at 4:10 pm

0
• May 15, 2020 at 7:55 pm

0
• April 26, 2020 at 9:08 am

Thanks you Alot .

0
• April 21, 2020 at 12:03 pm

0
• April 19, 2020 at 8:46 am