Freezing point of a $4 %$ aqueous solution of X is equal to freezing point of $12 %$ aqueous solution of Y.
If molecular weight of X is A, then molecular weight of Y is

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answer is B.

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Detailed Solution

For same freezing point, molality of both solution should be same.

$\begin{array}{l} m_{x} = m_{y} \\ \frac{4 \times 1000}{96 \times M_{x}} = \frac{12 \times 1000}{88 \times M_{y}} \\ or, M_{y} = \frac{96 \times 12}{4 \times 88} M_{x} = 3.27 A \end{array}$

Closest option is $3 A$ .

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