Solutions — JEE Main practice

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Sample questions with solutions

Q1 · 2026

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and $PQ_2$ . When 1 g of PQ is dissolved in 50 g of solvent 'A', $\Delta T_b$ was 1.176 K while when 1 g of $PQ_2$ is dissolved in 50 g of solvent 'A', $\Delta T_b$ was 0.689 K. ( $K_b$ of 'A' = 5 K kg mol⁻¹). The molar masses of elements P and Q (in g mol⁻¹) respectively, are :

A.
60, 25
B.
65, 145
C.
25, 60
D.
70, 110

Answer: C

For a non-volatile, non-ionizable solute, boiling point elevation is $\Delta T_b=K_bm$ , where molality $m$ depends on the solute's molar mass $M$ . With 1 g solute in 50 g (0.05 kg) solvent.

m = \frac{1/M}{0.05} = \frac{20}{M} \;\Rightarrow\; \Delta T_b = 5\times\frac{20}{M} = \frac{100}{M}

Rearranged, this gives the molar mass in terms of the observed elevation.

M = \frac{100}{\Delta T_b}

For PQ, substitute $\Delta T_b=1.176$ K.

M(PQ) = \frac{100}{1.176} \approx 85

Letting the molar masses of the elements be $x$ (for P) and $y$ (for Q):

x+y = 85 \quad (1)

For $PQ_2$ , substitute $\Delta T_b=0.689$ K.

M(PQ_2) = \frac{100}{0.689} \approx 145 \;\Rightarrow\; x+2y = 145 \quad (2)

Subtract equation (1) from (2) to isolate $y$ .

(x+2y)-(x+y) = 145-85 \;\Rightarrow\; y = 60

Substitute back into (1) to find $x$ .

x+60 = 85 \;\Rightarrow\; x = 25

Hence, the molar masses of P and Q are 25 and 60 g/mol respectively, so the answer is option C.

Q2 · 2026

Given below are two statements :

Statement I : The Henry's law constant $K_H$ is constant with respect to variations in solution's concentration over the range for which the solution is ideally dilute.

Statement II : $K_H$ does not differ for the same solute in different solvents.

In the light of the above statements, choose the correct answer from the options given below

A.
Both Statement I and Statement II are true
B.
Statement I is true but Statement II is false
C.
Statement I is false but Statement II is true
D.
Both Statement I and Statement II are false

Answer: B

For an ideally dilute solution, Henry's law states $p=K_Hx$ . At a fixed temperature, $K_H$ remains constant over the concentration range where the solution behaves ideally (dilute enough), confirming Statement I is true.
However, $K_H$ depends on both the nature of the gas (solute) and the nature of the solvent (as well as temperature). This means for the same solute, $K_H$ can differ across different solvents, making Statement II false. Hence, Statement I is true but Statement II is false, so the answer is option B.

Q3 · 2026

Consider a solution of $CO_2(g)$ dissolved in water in a closed container.

Which one of the following plots correctly represents variation of log (partial pressure of $CO_2$ in vapour phase above water) [y-axis] with log (mole fraction of $CO_2$ in water) [x-axis] at $25^{\circ}$ C?

Answer: C

Henry's Law states that the partial pressure of a gas above a solution is directly proportional to its mole fraction in the solution.

P(g) = K_H\cdot X(g)

Taking the logarithm of both sides converts this into a linear relationship.

\log P(g) = \log K_H+\log X(g)

Comparing this with the standard straight-line form $y=c+x$ , here $y=\log P(g)$ , $x=\log X(g)$ , and the intercept is $\log K_H$ .
This means the graph is a straight line with slope 1 and a y-intercept equal to $\log K_H$ . Hence, the correct graph shows this straight-line relationship with unit slope, so the answer is option C.

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