Mole Concept ‌–‌ ‌JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions with Solutions

Mole Concept ‌JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions‌ ‌with‌ ‌Solutions‌ ‌will help JEE aspirants develop better skills to solve different types of problems and prepare effectively for the upcoming entrance exam. We have listed down a set of previous year JEE Advanced questions that have been framed from the topic of the mole concept. Aspirants can therefore go through the questions and get an understanding of the question types, difficulty level and more. Additionally, the detailed solutions that we have provided will further help them to come up with the right approach to the answers.

Alternatively, as aspirants go through the Mole Concept JEE Advanced ‌Previous‌ ‌Year‌ ‌Questions‌ ‌with‌ ‌Solutions‌ ‌they will learn about the important topics to focus on from an exam perspective and study accordingly. To facilitate offline learning and practice we have also provided the questions and solutions in the form of a PDF. Students can download the PDF for free and use it to study productively at their own pace and time.

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JEE Advanced Previous Year Questions on Mole Concept

Question 1. In the chemical reaction between stoichiometric quantities of KMnO₄ and KI in a weakly basic solution, what is the number of moles of I₂ released for 4 moles of KMnO₄ consumed?

Solution: (6)

The chemical reaction of KMnO4 and KI in a weakly basic solution is given as;

Equivalents of KMnO₄ = Equivalents of I₂

n-factor × Number of moles (n) = n-factor × Number of moles (n)

3 × moles of KMnO4 = 2 × moles of I2

3 × 4 = 2 × moles of I₂

Moles of I₂ = 6 moles

Question 2. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl₂.6H₂O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952g of NiCl₂.6H₂O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is___. (Atomic weights in g mol–1: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

Solution: (2992)

Therefore, total mass = = 12 × 172 + 4 × 232 = 2992 g​

Question 3. Galena (an ore) is partially oxidized by passing air through it at high temperatures. After some time, the passage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O₂ consumed is ______. (Atomic weights in g mol^-1 : O = 16, S = 32, Pb = 207)

Solution: (6.47)

The given reaction will be:

2PbS + 3O₂ → 2PbO + 2SO₂

2PbO + PbS → 3Pb + SO₂

━━━━━━━━━━━━━━

3PbS + 3O₂ → 3Pb + 3SO₂

So, 3 mol of O₂ gives 3 mol of Pb

Thus, in other words, 96 kg of O₂ gives 621 kg of Pb.

So, 1 kg of O₂ will give 621 / 96 = 6.47kg

Question 4. If the value of the Avogadro number is 6.023 × 10²³ and the value of the Boltzmann constant is 1.380 × 10^-23 J/K, then the number of significant digits in the calculated value of the universal gas constant is.

Solution: (4)

We have, Universal gas constant,

R = kN_A

Where, k = Boltzmann constant and N_A = Avogadro’s number

Therefore, R = 1.380 × 10^-23 × 6.023 × 10²³ J/Kmol

= 8.31174 ≅ 8.312

Since k and N_A both have four significant figures, so the value of R is also rounded off up to 4 significant figures.

When the number is rounded off, the number of the significant figure is reduced.

So, the number of the significant figure is 4.

Question 5. Which has the maximum number of atoms?

A. 24 g of C (12)

B. 56 g of Fe (56)

C. 27 g of A1 (27)

D. 108 g of Ag (108)

Solution: (A)

Number of atoms = Number of moles

Avogadro’s number (N_A)

Number of atoms in 24 g C = 24 /12 × N_A = 2N_A

Number of atoms in 56 g of Fe = 56 / 56 N_A = N_A

Number of atoms in 27 g of Al = 27 / 27 N_A= N_A

Number of atoms in 108 g of Ag = 108 / 108 N_A = N_A

Hence, 24 g of carbon has the maximum number of atoms.

Question 6. How many moles of electrons weigh one kilogram?

A. 6.022 x 10²³

B. 1 / 9.108 x 10³¹

C. 6.022 / 9.108 x 10⁵⁴

D. 1 / 9.108 × 6.022 x 10⁸

Solution: (D)

Mass of one electron = 9.108 × 10^-31 kg.

Mass of one mole of electrons =9.108 × 10^-31 × 6.023 × 10²³ = 9.108 × 6.023 × 10^-8 kg.

The number of moles of electrons that weigh one kilogram

= 1 / 9.108 × 6.023 × 10^-8 = 1 / 9.108 × 6.023 × 10⁸

Hence, the correct option is D

Question 7. Calculate the molarity of water if its density is 1000 kg/m³.

Solution: (55.56)

Given, the density of the water = 1000 kg/m³

We know, 1 litre of water = 1 kg.

Since the density is 1000 kg/m³

Thus, 1000 ml of water = 1 kg.

(Since, 1 liter = 1000 ml)

Now, Mass of Water = 1 kg.

= 1000 g.

The molecular mass of the water molecules = 18 grams.

Using the formula,

No.of moles = mass / molar mass

= 1000 / 18

= 55.56 moles

For finding Molarity, we use the given formula,

Molarity = Number of moles / volumeinml × 1000

= 55.56 / 1000 × 1000

= 55.56 M

Question 8. Mixture X = 0.02 mole of [Co(NH₃)₅ SO₄] Br and 0.02 mole of [Co(NH₃)rBr]SO₄ was prepared in 2 L of solution.

1 L of the mixture if X + excess AgNO₃ → Y

1 L of mixture X+ excess BaCl₂ → Z

Number of moles of y and Z are:

A. 0.01, 0.01

B. 0.02, 0.01

C. 0.01, 0.02

D. 0.02, 0.02

Solution: (A)

In 1 L solution, there will be 0.01 mole of each [Co(NH₃)₅ SO₄] Br and [Co(NH₃)rBr]SO₄.

Addition of an excess of AgNO₃ will give 0.01 mole of AgBr.

Addition of an excess of BaCI₂ will give 0.01 mole of BaSO₄.

Question 9. An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

A. 40 mL

B. 20 mL

C. 10 mL

D. 4 ml

Solution: (A)

n-factor for H₂C₂O₄.2H₂O = 2

Normality = weight × 2 × 1000 / molecular weight × 250 = 6.3 × 2000 / 126 × 250 = 0.4

equating no. of equivalents:

N₁V₁ = N₂V₂

0.4 × 10 = V₂ × 0.1

Volume (V₂) = 40mL.

Hence, the correct option is A

Question 10. The normality of 0.3 M phosphorous acid H₃Po₃ is:

A. 0.1

B. 0.9

C. 0.3

D. 0.6

Solution: (D)

Normality = molarity × basicity (for acids)

Therefore, Normality = 0.3 × 2 = 0.6

∵ The basicity of H₃PO₃ is 2 because one of the H is not replaceable in H₃PO₃. The H atoms which are linked to oxygen are replaceable, but the H atom linked directly to central atom P is non-replaceable.

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