JEE Main Chemistry: Liquid Solution High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27)

Correct Answer: (a)

Explanation: Phenol + water and ethyl alcohol + water both forms homogeneous mixture but homogeneousily of phenol + water is more as compared to ethyl alcohol + water ( $\because$ larger alkyl group). Thus, the correct option is ' a '.

Correct Answer: (a)

Explanation: Ethanol + Acetone show positive deviation from Raoult's law while other solution show negative deviations.

Correct Answer: (b)

Explanation: Moles of glucose: $ n=\frac{18}{180}=0.1\ \text{mol} $ Mass of water: $ 100\ \text{g}=0.1\ \text{kg} $ Molality of the solution: $ m=\frac{0.1}{0.1}=1\ \text{mol kg}^{-1} $ Glucose does not dissociate in water, so the elevation in boiling point is: $ \Delta T_b=K_bm $ $ \Delta T_b=0.5\times1=0.5\ \text{K} $ Hence, the boiling point of the solution is: $ T_b=100+0.5=100.5^\circ\text{C} $ So, the correct answer is **Option B: $100.5^\circ\text{C}$**.

Correct Answer: (a)

Explanation: $\because$ Molar mass of glucose $(M)=180 \mathrm{~g}$ Molar mass of solute $(A)$, i.e. $\left(M_A\right)=58.5$ \% of $A=1.17 \%$ \% of glucose = 7.2\% Also, $\because \pi($ osmotic pressure $)=C R T$ $ \pi=\frac{n}{V} \cdot R T \Rightarrow \pi=\frac{W}{M} \times R T $ where, $\pi=$ osmotic pressure, $w=\%$ of sample $ \begin{aligned} M & =\text { molar mass, } R=\text { gas constant } \\ T & =\text { temperature } \end{aligned} $ ∵ These are isotonic. $ \begin{aligned} \therefore \quad & i=\frac{M_A \times \% \text { of glucose }}{\% \text { of } A \times M} \\ & i=\frac{58.5 \times 7.2}{1.17 \times 180}=\frac{421.2}{210.6}=2 \end{aligned} $ Hence, option (a) is the correct answer.

Correct Answer: (a)

Explanation: The relation between the vapor pressure of a solution, the vapor pressure of the pure solvent, and its mole fraction in the solution can be described by **Raoult's Law**. Raoult's Law states that the partial vapor pressure of a component in a solution is directly proportional to the mole fraction of the component in the solution and the vapor pressure of the pure component. Raoult's Law for a solvent in a solution can be expressed as: $ P_1 = P_1^\circ x_1 $ Where: $ P_1 $ is the vapor pressure of the solvent in the solution. $ P_1^\circ $ is the vapor pressure of the pure solvent. $ x_1 $ is the mole fraction of the solvent in the solution. Therefore, the correct expression according to Raoult's Law is given by Option A: $ \mathrm{P}_1=\mathrm{P}_1^{\circ} x_1 $