Question

Two solutions $A$ and $B$ are prepared by dissolving $1 g$ of non-volatile solutes $X$ and $Y$ respectively in $1 \,kg$ of water The ratio of depression in freezing points for $A$ and $B$ is found to be $1: 4$ The ratio of molar masses of $X$ and $Y$ is :

• A. $1: 4$
• B. $1: 0.25$
• C. $1: 0.20$
• D. $1: 5$

Answer 1

The Correct Option is B

The problem involves determining the ratio of molar masses of two non-volatile solutes, \(X\) and \(Y\), based on the depression in freezing points of their solutions. Let's solve it step-by-step.

Concept: The depression in freezing point (\(\Delta T_f\)) is calculated using the formula:

\(\Delta T_f = i \cdot K_f \cdot m\),

where:

• \(i\) is the van't Hoff factor (1 for non-electrolytes).
• \(K_f\) is the cryoscopic constant of the solvent (water in this case).
• \(m\) is the molality of the solution.

Since \(X\) and \(Y\) are non-electrolytes, \(i = 1\).

For a solution, molality \((m)\) is given by:

\(m = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}}.\)

Given:

• Mass of solute \(X = 1 \text{ g}\) and \(Y = 1 \text{ g}\).
• Mass of solvent = \(1 \text{ kg}\).
• Ratio of depression in freezing points: \(\Delta T_f(A) : \Delta T_f(B) = 1 : 4\).

Calculation:

Use the depression in freezing point formula:

\(\Delta T_f(A) = K_f \cdot \left(\frac{1}{M_X}\right)\) and \(\Delta T_f(B) = K_f \cdot \left(\frac{1}{M_Y}\right)\),

where \(M_X\) and \(M_Y\) are the molar masses of \(X\) and \(Y\) respectively.

Given that:

\(\frac{\Delta T_f(A)}{\Delta T_f(B)} = \frac{1}{4}\), substitute the expressions for \(\Delta T_f\):

\(\frac{K_f \cdot \left(\frac{1}{M_X}\right)}{K_f \cdot \left(\frac{1}{M_Y}\right)} = \frac{1}{4}\)

Simplify to find:

\(\frac{M_Y}{M_X} = 4\)

The ratio of molar masses is \(M_X : M_Y = 1 : 0.25\).

Therefore, the correct answer is \(1 : 0.25\). This corresponds to the option $1: 0.25$.