An aqueous solution of a non-electrolyte boil | vedclass

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(B) Given that the boiling point of the aqueous solution is $100.52 \, ^\circ C$,the elevation in boiling point is $\Delta T_b = 100.52 \, ^\circ C -

Vedclass · Accepted Answer

$- 1.86$

Vedclass · Answer

(B) Given that the boiling point of the aqueous solution is $100.52 \, ^\circ C$,the elevation in boiling point is $\Delta T_b = 100.52 \, ^\circ C - 100 \, ^\circ C = 0.52 \, ^\circ C$.For water,the molal elevation constant $K_b = 0.52 \, K \cdot kg \cdot mol^{-1}$.Using the formula $\Delta T_b = K_b 	imes m$,we get $0.52 = 0.52 	imes m$,which implies the molality $m = 1 \, mol \cdot kg^{-1}$.The depression in freezing point is given by $\Delta T_f = K_f 	imes m$.For water,the molal depression constant $K_f = 1.86 \, K \cdot kg \cdot mol^{-1}$.Thus,$\Delta T_f = 1.86 	imes 1 = 1.86 \, ^\circ C$.The freezing point of the solution $T_f = 0 \, ^\circ C - \Delta T_f = 0 \, ^\circ C - 1.86 \, ^\circ C = - 1.86 \, ^\circ C$.

List-$I$	List-$II$
$A$. van't Hoff factor,$i$	$I$. Cryoscopic constant
$B$. $k_{f}$	$II$. Isotonic solutions
$C$. Solutions with same osmotic pressure	$III$. $\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$
$D$. Azeotropes	$IV$. Solutions with same composition of vapour above it

An aqueous solution of a non-electrolyte boils at $100.52 \, ^\circ C$. The freezing point of the solution will be ............ $^\circ C$.

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Distilled water boils at $373.15 \ K$ and freezes at $273.15 \ K$. $A$ solution of glucose in distilled water boils at $373.202 \ K$. What is the freezing point (in $K$) of the same solution? (For water, $K_{b}=0.52 \ K \ kg \ mol^{-1}, \ K_{f}=1.86 \ K \ kg \ mol^{-1}$)

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