Freezing point of an aqueous solution is $(-0.186)^{\circ} \mathrm{C}$. Elevation of boiling point of the…

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.
  1. $0.186{ }^{\circ} \mathrm{C}$
  2. $0.0512{ }^{\circ} \mathrm{C}$
  3. $0.092{ }^{\circ} \mathrm{C}$
  4. $0.2372{ }^{\circ} \mathrm{C}$

Solution

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

Asked in: JEE Main 2002

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