Freezing point of an aqueous solution is $-0.186^{\circ} \mathrm{C}$. If the values of $\mathrm{K}_{b}$ and…

Freezing point of an aqueous solution is $-0.186^{\circ} \mathrm{C}$. If the values of $\mathrm{K}_{b}$ and $\mathrm{K}_{f}$ of water are respectively $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ and $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, then the elevation of boiling point of the solution in $\mathrm{K}$ is
  1. $0.52$
  2. $1.04$
  3. $1.34$
  4. $0.052$

Solution

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \cdot \mathrm{K}_{\mathrm{f}} \cdot \mathrm{m} ; \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} . \mathrm{K}_{\mathrm{b}} \cdot \mathrm{m}$
$\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{b}}}=\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}$
$\Delta \mathrm{T}_{\mathrm{f}}=0-\left(-0.186^{\circ} \mathrm{C}ight)=0.186^{\circ} \mathrm{C}$
$\frac{0.186}{\Delta \mathrm{T}_{\mathrm{b}}}=\frac{1.86}{0.52} \Rightarrow \Delta \mathrm{T}_{\mathrm{b}}=\frac{0.52 \times 0.186}{1.86}=0.052$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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