Q13 / 180 · KCET 2022 · Chemistry
The rise in boiling point of a solution containing $1.8 \mathrm{~g}$ of glucose in $100 \mathrm{~g}$ of solvent is $0.1^{\circ} \mathrm{C}$. The molal elevation constant of the liquid is
- Correct answerA$1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}$
- B$2 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}$
- C$10 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}$
- D$0.1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}$
Solution
Given, amount of glucose $(w)=1.8 \mathrm{~g}$
Amount of solvent $(W)=100 \mathrm{~g}$
$\Delta T_b=0.1^{\circ} \mathrm{C}$
Molecular mass of glucose $=180$
Molal elevation constant of the liquid, $\Delta T_b=0.1^{\circ} \mathrm{C}$
We know that,
$$
\begin{aligned}
K_b & =\frac{\Delta T_b \times m \times W}{1000 \times w} \\
& =\frac{0.1 \times 180 \times 100}{1000 \times 1.8}=1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.