Question

The boiling point of an aqueous solution of a non volatile solute is $100.15^{\circ} C .$ What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? The values of $K_{h}$ and $K_{f}$ for water are 0.512 and $1.86 K$ molality $^{-1}$

• A. $-0.544^{\circ} C$
• B. $-0.512^{\circ} C$
• C. $-0.272^{\circ} C$
• D. $-1.86^{\circ} C$

Answer 1

Answer: (C)

$\Delta T _{ b }= k _{ b } \times M...(1)$

If solution is diluted with equal volume of water, th concentration of solutions is $\frac{ M }{2}$

So, $\Delta T _{ f }= K _{ f } \times \frac{ M }{2}...(2)$

$\frac{(1)}{(2)}: \frac{\Delta T_{b}}{\Delta T_{b}}=\frac{K_{b} \times M}{K_{f} \times \frac{M}{2}}$

$=\frac{0.15}{\Delta T _{ f }}=\frac{0.512 \times 2}{1.86}$

$\Delta T _{ f }=0.272$

$\Delta T _{ f }=( F.P. )_{\text {solution }}-( F.P. )_{\text {solution }}$

$0.272=0^{\circ} C -( F.P. )_{\text {solution }}$

$(F.P.)_{\text{solution}} =-0.272^{\circ} C$