Question

Freezing point of an aqueous solution is $- 0.186\,{}^{\circ}\,C$. If the values of $K_b$ and $K_f$ of water are respectively $0.52\, K \,kg$ $mol^{-1}$ and $1.86 \,K\, kg\, mol^{-1}$, then the elevation of boiling point of the solution in $K$ is

• A. 0.52
• B. 1.04
• C. 1.34
• D. 0.134
• E. 0.052

Answer 1

Answer: (E)

$\because$ Freezing point of the aqueous solution

$=-0.186^{\circ} C$

$\therefore $ Depression in freezing point,

$\Delta T_{f} =0^{\circ} C -\left(-0.186^{\circ} C \right) $

$=+0.186^{\circ} C =+0.186 K$

We know that

$\Delta T_{f}=K_{f} \cdot m\,....(i)$

and elevation in boiling point,

$\Delta T_{b}=K_{b} \cdot m\,.....(ii)$

Eq. (i) Eq. (ii) gives

$\frac{\Delta T_{f}}{\Delta T_{b}} =\frac{K_{f}}{K_{b}} $

$ \Rightarrow \frac{0.186 K }{\Delta T_{b}} =\frac{1.86 \,K \,kg \,mol ^{-1}}{0.52\,K\,kg\, mol ^{-1}} $

$ \Delta T_{b} =\frac{0.186 \times 0.52}{1.86} $

$ =0.052 \,K $