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Mole Concept - JEE Advanced Previous Year Questions with Solutions

Practice JEE Advanced Mole Concept questions covering atomic mass calculation, molarity, molality, density-based problems, mole fraction, and concentration relationships to strengthen numerical problem-solving skills in physical chemistry.

Mole Concept - JEE Advanced Previous Year Questions with Solutions

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Why Mole Concept Matters in JEE Advanced 

The mole concept is the foundation of physical chemistry. In JEE Advanced, it appears both as a standalone topic and as the entry point for questions on solutions, electrochemistry, and chemical equilibrium. According to the NTA's official JEE Advanced syllabi, mole concept falls under Physical Chemistry – Basic Concepts, a section that has contributed 1–3 questions in nearly every JEE Advanced paper since 2009.

What makes these questions tricky is not the formula — it is the unit conversion. Students who can mechanically apply the molarity formula still lose marks because they miss a density conversion from g/cm³ to kg/L, or confuse volume of solution with volume of solvent.

Mole Concept - 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.

Q. Given that the abundances of isotopes $^{54} \mathrm{Fe},^{56}$ Fe and $^{57}$ Fe are $5 \%, 90 \%$ and $5 \%$, respectively, the atomic mass of Fe is : (A) 55.85               (B) 55.95                (C) 55.75               (D) 56.05 [JEE 2009]
Ans. (B) $\mathrm{M}_{\mathrm{AVg}}=\frac{\mathrm{M}_{1} \mathrm{X}_{1}+\mathrm{M}_{2} \times \mathrm{X}_{2}+\mathrm{M}_{3} \times \mathrm{X}_{3}}{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}$ $=\frac{5 \times 54+56 \times 90+5 \times 57}{5+90+5}$ = 55.95
Q. Silver (atomic weight $=108 \mathrm{g} \mathrm{mol}^{-1}$ ) has a density of 10.5 g $\mathrm{cm}^{-3}$. The number of silver atoms on a surface of area $10^{-12} \mathrm{m}^{2}$ can be expressed in scientific notation as y ´ $10^{\mathrm{x}}$. The value of x is - [JEE 2010]
Ans. 7 $\mathrm{d}=\frac{\text { mass }}{\mathrm{V}} \Rightarrow 10.5 \mathrm{g} / \mathrm{cm}^{3}$ means in $1 \mathrm{cm}^{3}=10.5 \mathrm{g}$ of $\mathrm{Ag}$ No of atom of $\mathrm{Ag}$ in $1 \mathrm{cm}^{3}=\frac{10.5}{108} \times \mathrm{N}_{\mathrm{A}}$ in $1 \mathrm{cm}, \mathrm{no}$ of atom of $\mathrm{Ag}=\left(\frac{10.5}{108} \times \mathrm{N}_{\mathrm{A}}\right)^{1 / 3}$ in $10^{-12} \mathrm{m}^{2}$ or $10^{-8} \mathrm{cm}^{2},$ No of atom of $\mathrm{Ag}$ $=\left(\frac{10.5 \mathrm{N}_{\mathrm{A}}}{108}\right)^{2 / 3} \times 10^{-8}=\left(\frac{1.05 \times 6.022 \times 10^{23}}{108}\right)^{2 / 3} \times 10^{-8}$ $=1.5 \times 10^{7}$ Hence $\mathrm{x}=7$
Q. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is (A) 1.78 M          (B) 2.00 M             (C) 2.05 M              (D) 2.22 M [JEE 2011]
Ans. (C) Vol of Solution $=\frac{\text { mass }}{\text { density }}$ $=\frac{1000+120}{1.15} \mathrm{mL}$ $=\frac{1120}{1.15} \mathrm{ml}$ Molarity $=\frac{120 / 60}{\frac{1120}{1.15}} \times 1000=2.05 \mathrm{M}$
Q. A compound $\mathrm{H}_{2} \mathrm{X}$ with molar weight of 80 g is dissolved in a solvent having density of 0.4 g /ml, Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is. [JEE 2014]
Ans. 8 $\mathrm{V}_{\text {solvent }}=\mathrm{V}_{\text {solution }}$ 3.2 mol present in 1L solution (solvent) 3.2 mol / solvent $1000 \times 0.4=400 \mathrm{gm}$ $=.4 \mathrm{Kg}$ Molality $=\frac{3.2}{0.4}=8 \mathrm{m}$
Q. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g $\mathrm{cm}^{-3}$. The ratio of the molecular weights of the solute and solvent, $\left(\frac{\mathrm{MW}_{\text {solute }}}{\mathrm{MW}_{\text {solikent }}}\right)$, is [JEE - ADV. 2016]
Ans. 9 1 mole solution has 0.1 mole solute and 0.9 mole solvent Let $\mathrm{M}_{1}=$ Molar mass solute $\mathrm{M}_{2}=$ Molar mass solvent

Frequently Asked Questions

Find answers to common questions.

How many questions come from the mole concept in JEE Advanced?

Typically 1–2 questions appear from mole concept in JEE Advanced each year, usually in Paper 1. They are often integer-type or single-correct MCQs. While the weightage is not the highest, these questions are considered scoring because the formulas are fixed and errors are avoidable with careful unit management.

Is mole concept important for JEE Advanced or only JEE Main?

Mole concept is tested in both JEE Main and JEE Advanced. In JEE Advanced, the questions are more application-based — for example, combining density, Avogadro's number, and surface area in a single problem (as seen in the 2010 silver atoms question). It also acts as a prerequisite for electrochemistry and solution chemistry at the JEE Advanced level.

What is the fastest way to calculate average atomic mass in JEE?

Use the weighted average formula directly: M_avg = Σ(mass of isotope × abundance%) ÷ 100. Do not divide by the number of isotopes. For the 2009 Fe question, this gives (54×5 + 56×90 + 57×5) ÷ 100 = 55.95. This formula works for any number of isotopes and takes under 30 seconds with practice.

What is the difference between molarity and molality in JEE problems?

Molarity (M) = moles of solute per litre of solution — it changes with temperature because volume changes. Molality (m) = moles of solute per kilogram of solvent — it is temperature-independent. In JEE Advanced 2016, the question exploited the condition where M = m at a specific density and mole fraction to find a molecular weight ratio.

Are there mole concept questions in JEE Advanced after 2016?

Yes. While this page currently covers 2009–2016, mole concept continues to appear post-2016, often embedded within thermochemistry, solutions, or electrochemistry questions rather than as a standalone topic. Checking the full JEE Advanced question papers year-wise on eSaral will give you the most current coverage.

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