1. Use the formula for depression in freezing point

For a dilute solution of a non-volatile solute,

$\Delta T_f = i K_f m$

Here, since both solutes are non-volatile and no dissociation is mentioned, we take

$i=1$

So,

$\Delta T_f \propto m$

where molality,

$m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

2. Write molality for both solutions

Each solution contains $1\,\text{g}$ solute dissolved in $1\,\text{kg}$ water.

For solution A (solute $X$ of molar mass $M_X$):

$m_A = \frac{1/M_X}{1} = \frac{1}{M_X}$

For solution B (solute $Y$ of molar mass $M_Y$):

$m_B = \frac{1/M_Y}{1} = \frac{1}{M_Y}$

3. Use the given ratio of depression in freezing points

Given,

$\Delta T_{f,A} : \Delta T_{f,B} = 1:4$

Since $\Delta T_f \propto m$,

$m_A : m_B = 1:4$

Substitute molalities:

$\frac{1}{M_X} : \frac{1}{M_Y} = 1:4$

This means

$\frac{1/M_X}{1/M_Y} = \frac{1}{4}$

$\frac{M_Y}{M_X} = \frac{1}{4}$

$\frac{M_X}{M_Y} = 4$

So,

$M_X : M_Y = 4:1 = 1:0.25$

4. Check options

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

Therefore, the correct answer is Option B.