Q21 / 160 · AP EAMCET 2023 (18 May Shift 1) · Chemistry
What is the boiling point of solution of $0.1 \mathrm{~m} \mathrm{KCl}$ ? $\mathrm{K}_{\mathrm{b}}$ of water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \cdot(\alpha=100 \%)$(water boil at $373 \mathrm{~K}$ )
- A$100.104 \mathrm{~K}$
- Correct answerB$373.104 \mathrm{~K}$
- C$273.104 \mathrm{~K}$
- D$373.052 \mathrm{~K}$
Solution
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \times \mathrm{i}=(0.52)(0.1)(2)=0.104 \mathrm{~K}$
$\Rightarrow \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}^0+\Delta \mathrm{T}_{\mathrm{b}}=373+0.104=373.104 \mathrm{~K}$
Hence, the correct answer is option: 'B'.