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Specific Heat Capacity

Freezing poin...

Question

Freezing point of a $5 %$ aqueous solution of $X$ is equal to freezing point of $20 %$ aqueous solution of $Y$ . If molecular weight of $X$ is $A$ , then molecular weight of $Y$ is:

4.75A

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Solution

The correct option is A 4.75A
For same freezing point, molality of both solution should be same.
$m_{x} = m_{y}$
$m_{x} = \frac{weight of x \times 1000}{molecular weight (Mwt) of \times (M_{x}) \times weight of solvent}$
$m_{y} = \frac{weight of y \times 1000}{M_{y} \times weight of solvent}$
So, $\frac{5 \times 1000}{95 \times M_{x}} = \frac{20 \times 1000}{80 \times M_{y}}$ or,
$M_{y} = \frac{95 \times 20}{5 \times 80} M_{x} = 4.75$ A
Closest option is 4.75A.

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