Q51 / 200 · WBJEE 2014 · Chemistry
To observe an elevation of boiling point of $0.05^{\circ} \mathrm{C},$ the amount of a solute (mol. wt. $=100$ ) to be added to $100 \mathrm{g}$ of water $\left(K_{b}=0.5\right)$ is
- A$2 \mathrm{g}$
- B$0.5 \mathrm{g}$
- Correct answerC$1 \mathrm{g}$
- D$0.75 \mathrm{g}$
Solution
Elevation of boiling point,
$$
\Delta T_{b}=\frac{w \times K_{b} \times 1000}{M \times W}
$$
(Here, w and $W=$ weights of solute and solvent respectively.
$M=$ molecular weight of solute and $K_{b}=$ constant
On substituting values, we get
of
$$
\begin{aligned}
0.05 &=\frac{W \times 0.5 \times 1000}{100 \times 100} \\
w &=\frac{0.05 \times 100 \times 100}{0.5 \times 1000}=1 \mathrm{g}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.