Question

Two solutions containing $0.75\, g$ of urea (mol. wt. 60) and $1.5 \,g$ of compound $A$ in $100\, g$ water, freeze at the same temperature. The molecular weight of $A$ is:

• A. 60
• B. 30
• C. 120
• D. 240

Answer 1

Answer: (C)

Calculate $\Delta T_{f}$ for both the solutions (containing urea and $A$ ) separately divide the two equations to get the molecular mass of $A$.
Given mass of urea $=0.75\, g$, mass of $=1.5\, g$,
mass of water $=100\, g$, mol. $wt$, of urea $=60 .$
$\Delta T_{f}$ for urea
$=\frac{1000 \times K_{f} \times \text { mass of urea }}{\text { mass of water } \times \text { molecular mass of urea }}$
$=\frac{1000 \times K_{f} \times 0.75}{60 \times 100}\,\,\,...(i)$
$\Delta T_{f}$ for ' $A$ '
$=\frac{1000 \times K_{f} \times \text { mass of } A}{\text { mass of water } \times \text { molecular mass of } A}$
$=\frac{1000 \times K_{f} \times 1.5}{100 \times \text { molecular mass of } A}\,\,\,\,...(ii)$
Dividing Eq. (i) by (ii), we get
$\frac{1000 \times K_{f} \times 0.75 \times 100 \times \text { molar mass of } A}{1000 \times K_{f} \times 1.5 \times 60 \times 100}$
$=\frac{0.75 \times \text { molecular mass of } A}{1.5 \times 60}$
Molecular mass of $A=120$