Q23 / 160 · TS EAMCET 2018 (04 May Shift 2) · Chemistry
Calculate the approximate $\Delta T_b$ (in K) for 0.001 molal $\mathrm{KCl}$ solution, if its van't-Hoff factor is $1.98\left[K_b\right.$ of water is $\left.0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$
- A1.03
- Correct answerB$1.03 \times 10^{-3}$
- C$1.03 \times 10^{-5}$
- D$1.03 \times 10^{-1}$
Solution
We know that
$$
\Delta T_b=i K_b m \quad\left(\because \Delta T_b=i . K_b \times m\right)
$$
where, $\Delta T_b=$ elivation in boiling point (to time)
$$
\begin{aligned}
i & =\text { van't-Hoff factor }(1.98) \\
K_b & =\text { molal elevation constant }(0.52) \\
m & =\text { molality }(0.001 \mathrm{~m}) \\
\Delta T_b & =1.98 \times 0.52 \times 0.001=0.0010296 \\
\Delta T_b & =1.03 \times 10^{-3}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.