Q5 / 170 · Manipal (MET) 2019 · Chemistry
An aqueous solution freezes at $-0.186^{\circ} \mathrm{C}$, then elevation in boiling point is
$\left(K_b=0.512, K_f=1.86\right)$
- Correct answerA$0 . 0512^{\circ} \mathrm{C}$
- B$100.0512^{\circ} \mathrm{C}$
- C$-0.0512^{\circ} \mathrm{C}$
- DNone of these
Solution
Elevation in boiling point $\Delta T_b=K_b \times m$
Dipression in freezing point $\begin{aligned} & \Delta T_f=K_f \times m \\ & \frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f} \end{aligned}$ $\begin{aligned} \frac{\Delta T_b}{0.186} & =\frac{0.512}{1.86} \\ \Delta T_b & =0.0512^{\circ} \mathrm{C}\end{aligned}$
Dipression in freezing point $\begin{aligned} & \Delta T_f=K_f \times m \\ & \frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f} \end{aligned}$ $\begin{aligned} \frac{\Delta T_b}{0.186} & =\frac{0.512}{1.86} \\ \Delta T_b & =0.0512^{\circ} \mathrm{C}\end{aligned}$
Hence, the correct answer is option: 'A'.