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 -

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$- 1.86$

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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 $\Delta T_b = 100.52 \, ^\circ C - 100 \, ^\circ C = 0.52 \, ^\circ C$.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 $T_f = 0 \, ^\circ C - \Delta T_f = 0 \, ^\circ C - 1.86 \, ^\circ C = - 1.86 \, ^\circ C$.

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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