Q22 / 160 · AP EAMCET 2024 (20 May Shift 2) · Chemistry
A non-volatile solute is dissolved in water. The $\Delta T_b$ of resultant solution is 0.052 K . What is the freezing point of the solution (in K)?
( $\mathrm{K}_{\mathrm{b}}$ of water $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ; \mathrm{K}_{\mathrm{f}}$ of water $=1.86 \mathrm{~K} \mathrm{~kg}$ $\mathrm{mol}^{-1}$, Freezing point of water $=273 \mathrm{~K}$ )
- A$272.628$
- B$273.186$
- C$273.000$
- Correct answerD$272.814$
Solution
Given
$\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=0.052 \mathrm{~K} \\
& 0.052 \mathrm{~K}=0.52 \times \mathrm{m} \\
& \mathrm{~m}=0.1 \\
& \Delta \mathrm{~T}_{\mathrm{f}}=\mathrm{m} \mathrm{~K}_{\mathrm{f}}=0.1 \times 1.86=0.186
\end{aligned}$
Freezing Point of solution $=273-0.186=272.81 \mathrm{~K}$.
Freezing Point of solution $=273-0.186=272.81 \mathrm{~K}$.
Hence, the correct answer is option: 'D'.