Q21 / 160 · AP EAMCET 2024 (19 May Shift 2) · Chemistry

The $\Delta \mathrm{T}_{\mathrm{b}}$ value for 0.01 m KCI solution is 0.01 K . What is the Van't Hoff factor? $\left(\mathrm{K}_{\mathrm{b}} \text { for water }=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$

  • Correct answer
    A
    1.92
  • B
    1.72
  • C
    0.96
  • D
    0.86

Solution

Given, $\Delta \mathrm{T}_{\mathrm{b}}=0.01 \mathrm{~K}$ Molality $=0.01 \mathrm{~m}$ $\mathrm{K}_{\mathrm{b}}=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ We know, $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{K}_{\mathrm{b}} \mathrm{m}$ $\mathrm{i}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{~K}_{\mathrm{b}} \mathrm{~m}}=\frac{0.01}{0.52 \times 0.01}=1.92$

Hence, the correct answer is option: 'A'.

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