Question

Freezing point of a $5\%\left(w/w\right)$ aqueous solution of $X$ is equal to freezing point of $20\%\left(w/w\right)$ aqueous solution of $Y$. If molecular weight of $X$ is $A$, then molecular weight of $Y$ is:

• A. $4.75A$
• B. $1.75A$
• C. $4A$
• D. $2.5A$

Answer 1

The correct option is A $4.75A$

Since we are considering only aqueous solution here, thus for same freezing point, molality of both solution should be same.
$m_X=m_Y$
$m_X=\frac{\text{Weight of X (in g)}\times 1000}{\text{Molecular weight of X}\left(M_X\right)\times \text{Weight of solvent (in g)}}=\frac{5\times 1000}{95\times M_X}$
$m_Y=\frac{\text{Weight of Y (in g)}\times 1000}{\text{Molecular weight of Y}\left(M_Y\right)\times \text{Weight of solvent (in g)}}=\frac{20\times 1000}{80\times M_Y}$
So,
$\frac{5\times 1000}{95\times M_X}=\frac{20\times 1000}{80\times M_Y}$
$\text{Given,}{M}_X=A$
$\Rightarrow M_Y=\frac{95\times 20}{5\times 80}{M}_X=4.75A$