Q39 / 225 · JEE Main 2002 · Chemistry

Freezing point of an aqueous solution is $(-0.186)^{\circ} \mathrm{C}$. Elevation of boiling point of the same solution is $\mathrm{K}_{\mathrm{b}}=0.512^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{f}}=1.86^{\circ} \mathrm{C}$, find the increase in boiling point.

  • A
    $0.186{ }^{\circ} \mathrm{C}$
  • Correct answer
    B
    $0.0512{ }^{\circ} \mathrm{C}$
  • C
    $0.092{ }^{\circ} \mathrm{C}$
  • D
    $0.2372{ }^{\circ} \mathrm{C}$

Solution

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}} \times 1000 ; \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}} \times 1000 ; \frac{\Delta \mathrm{T}_{\mathrm{b}}}{\Delta \mathrm{T}_{\mathrm{f}}}=\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{-0.186}=\frac{0.512}{1.86}=0.0512^0 \mathrm{C}$

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

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