INTRODUCTION

In my blog post, Mole Concept IIT JEE PYQs with Solutions, I have tried to elaborate the concept with reference to questions. A mole is defined as the amount of a substance that contains exactly 6.02214076 × 10^23 ‘elementary entities’ of the given substance. The number 6.02214076× 10^23 entities (atoms/molecules/ions etc) is known as the Avogadro Number which is denoted by the symbol ‘NA’.

Mole Concept IIT JEE PYQs with Solutions

Given below are the questions related to mole concept which were asked in IIT JEE MAIN EXAMINATIONS in past years.

NOTE- Solutions are given at the end so that students can first practise the questions on their own and then they can check their answers.

1.

50 mL of acetic acid solution (0.06N) was added to 3g of activated charcoal in a flask. It was filtered  after an hour 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

 

2.

An open vessel at 300 K is heated till ⅖ th of the air in it is expelled. If 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

3.

In a particular gaseous mixture, the ratio of masses of oxygen and nitrogen is 1 : 4. The number of their molecule are in the ratio of  :

(1) 1 : 8

(2) 3 : 16

(3) 1 : 4

(4) 7 : 32

4.

What will be the composition of the gaseous mixture at STP ?

Upon passing 0.5 litre of CO2 (g) over red hot coke, the total volume of the gases increased to 700 ml. When CO2 (g) is passed over red hot coke it partially gets reduced to CO(g).(1) CO2 = 200 mL: CO = 500mL

(2) CO2 : 350 mL &  CO : 350mL

(3) CO2 : 0.0 mL& CO : 700mL

(4) CO2: 300 mL & CO : 400mL

5.

If the density of the concentrated sulphuric acid (peddled commercially is 95% H2SO4 by weight) is 1.834 g cm-3, what will be the molarity of this solution –

(1) 17.8 M

(2) 15.7 M

(3) 10.5 M

(4) 12.0 M

6.

 The density of a solution prepared by dissolving 120 g of urea in 1000 g of water is 1.15 g/mL.What will be the molarity of this solution? (mol. mass of urea= 60 u)

(1) 2.05 M

(2) 0.50 M

(3) 1.78 M

(4) 1.02 M

7.

One molecule of the compound ( CxHyOZ ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. Write the empirical formula of compound CxHyOz, if the ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1. 

(1) C2H4O

(2) C3H4O2

(3) C2H4O3

(4) C3H6O3

8.

Which of the following has the largest number of atoms in the following samples of substances:

(1) 127.0g of iodine

(2) 48.0g of magnesium

(3) 71.0g of chlorine

(4) 4.0g of hydrogen

9.

What will be the molality of the solution having the density of 3M solution of sodium chloride is 1.252 g mL-1.(molar mass, NaCl = 58.5 g mol-1)

(1) 2.18 m

(2) 3.00 m

(3) 2.60 m

(4) 2.79 m

10.

 When 0.6 g of urea is heated strongly with NaOH , NH3 is liberated.  Liberated NH3 will combine with which of the following HCl solution?

(1) 100 mL of 0.2 N HCl

(2) 400 mL of 0.2 N HCl

(3) 100 mL of 0.1 N HCl

(4) 200 mL of 0.2 N HCl

11.

16. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied.What can you say about the mole fraction of methyl alcohol in the solution?

(1) 0.086

(2) 0.050

(3) 0.100

(4) 0.190

12.

200 mL of 0.5 (M) of NaOH solution is added to 10 mL of 2(M) NaOH solution. What is the final concentration?

(1) 0.57 M

(2) 5.7 M

(3) 11.4 M

(4) 1.14 M

13.

 M is a  transition metal which forms a volatile chloride having vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be

(1) MCl2

(2) MCl4

(3) MCl5

(4) MCl3

14.

We have a solution of HNO3 having density 1.4 g/mL and 63% w/w. Determine the molarity of HNO3 solution.

15.

Calculate the mass of FeSO4.7H2O which must be added in 100 kg of wheat to get 10 PPM of Fe.

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Solutions:

1.

Molarity of CH3COOH solution = mass of acetic acid/molar mass)/volume of solution in litre

Acetic acid is monobasic.

0.042 = W/(60×0.05)

W = 0.042×60×0.05 = 0.126 g

Amount of acetic acid actually adsorbed = 0.180-0.126 = 54mg

Amount of charcoal available = 3 g

So amount of acetic acid adsorbed per gram of charcoal = 54mg×1g/3.0g = 18 mg

Hence option (3) is correct.

2.

At constant V and P, n1T1 = n2T2

n1 = n

n2 = n-2n/5 = 3n/5

T1 = 300 K

300 n = (3n/5) T2

T2 = 300×5/3 = 500 K

Hence option (3) is correct

3.

Given,

 Mass of oxygen:  Mass of nitrogen = 1:4

Let mass of O2 = w

Mass of N2 = 4w

Molecules of O2 = w/(32×NA)

Molecules of N2 = 4w/(28×NA)

Ratio of number of molecules = w/(32×NA)÷4w/(28×NA)

= w/(32×NA)×(28×NA)/4w

= 7/32

So the ratio is 7:32.

Hence option (4) is correct.

4.

CO2 (g) + C (s) → 2CO (g)

Total volume = 

                       700 ml

                      = 0.7 L

0.5+x = 0.7

x = .2L = 200 mL

CO2 (g) = 0.5-0.2 = 300ml

CO (g) = 2x = 400 mL

Hence option (4) is correct.

5.

Given Density = 1.834

1 ml solution contains 1.834 g

1000 ml solution will contain 1834 g

95% H2SO4 means 100 gm contain 95 gm H2SO4

Mass of solute = (95/100)×1834

Molecular weight of H2SO4 = 98

Molarity = No. of moles per unit volume 

             = mass of solute/98

             = (95/100)×(1834/98)

             = 17.8 M

Hence option (1) is correct.

6.

Given density of solution = 1.15g/mL

mass of solution = 1000+120 = 1120 gm

Molar mass = 60

Volume = mass /density of solution

= 1120/1.15

No. of moles = 120/60 = 2

Molarity = No. of moles/ volume

= 2÷ (1120×10-3/1.15)

= 2×1.15×1000 /1120)

= 2.05 M

Hence option (1) is correct.

7.

Given the ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1.

Atomic mass of carbon = 12

Atomic mass of Hydrogen = 1

If we have x atoms of Carbon and y atoms of Hydrogen,

12*x = 6(1*y)

12x = 6y

So y = 2x

Given one molecule of compound ( CxHyOZ ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O

CXHy + O2 → xCO2+ (y/2)H2O

Put y = 2x in above equation

CXH2x + O2 → xCO2+ xH2O

Oxygen needed = 2x+x = 3x

So, z gives half of oxygen required to burn.

So z = 3x/2 = 1.5 x

Check the given options which satisfies z = 1.5x.

So the empirical formula is C2H4O3.

Hence option (3) is correct.

8.

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

127.0g of iodine.

                           = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

                       = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

                                     = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

6. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

1. of atoms of hydrogen= 4 moles of hydrogen atom.

Hence option(4) is correct.

9.

Given Molar mass of NaCl = 58.5 g

M = 3 mol L-1

Mass of NaCl in 1L solution ,W2 = 3×58.5 = 175.5g

Mass of L solution = V × d

= 1000 ×1.25 = 1250g

Mass of H2O in solution (W1) = 1250-175.5 = 1074g

m = W2×1000/Mw2 ×W1 = (175.5×1000)/58.5×1074.5 = 2.79m

Hence option (4) is correct.

10.

NH2CONH2 + 2NaOH → Na2CO3 + 2NH3

2 mole of urea ≡ one mole of NH3

one mole of NH3 = one mole of HCl

So one mole of HCl = 2 mole of urea = 2×0.6/60 = 0.02 mol.

Hence option (1) is correct.

11.

We know mole fraction = moles of solute/(moles of solute + moles of solvent)

Let mass of water is 1 kg . Moles of CH3OH is 5.2

Xsolute = 5.2/(5.2+1000/18) = 5.2/(5.2+55.556) = 5.2/60.756 = 0.086

Hence option (1) is correct.

12.

No. of moles of NaOH in 10 mL of 2 M solution = (10/1000)×2 = 0.02 mol

Number of moles of NaOH in 200 mL of 0.5M solution = (200/1000)×0.5 = 0.1 mol

Total number of moles of NaOH = 0.02+0.1 = 0.12 mol

Total volume = 10+200 = 210 mL = 0.210 L

Final concentration = 0.12/0.210 = 0.57 M

Hence option (1) is correct.

13.

Given vapour density = 94.8

Vapour density = molecular mass/2

Molecular mass = 94.8×2 = 189.6

Given 74.75% chlorine.

So,  chloride = 74.75/100 * 189.6 = 141.72 g

Hence,

number of atoms of chloride =141.72/35.5 =3.97 (approximately 4).

So the metal chloride is MCl4.

Hence option (2) is correct.

14.

Density = mass/volume of solution

Volume =

mass / density

= (100/1.4)ml

Molarity = no. of moles of solute /Volume of solution( l) = 1.4×1000/100 = 14 M

Hence 14 M is correct.

15.

Ppm = (Mass of Fe/total mass)×106

Total mass = 100 kg = 100 × 1000 g

Mass of Fe = (ppm × total mass )/106

= 10× 100 × 1000/106 = 1 g

Molecular mass of FeSO4.7H2O = 278

Mass of one Fe = 56 g

56 g of Fe → 278 g of FeSO4.7H2O

So 1 g of Fe → 278/56 = 4.96 g

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