The molal elevation constant of water is 0.52 | vedclass

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(B) For $KCl$,the van't Hoff factor $i = 2$ because it dissociates as $KCl \rightarrow K^+ + Cl^-$.The formula for elevation in boiling point is $\Del

Vedclass · Accepted Answer

$101.04$

Vedclass · Answer

(B) For $KCl$,the van't Hoff factor $i = 2$ because it dissociates as $KCl ightarrow K^+ + Cl^-$.The formula for elevation in boiling point is $\Delta T_b = i 	imes K_b 	imes m$.Given $i = 2$,$K_b = 0.52 \, ^{\circ}C \, kg \, mol^{-1}$,and $m = 1.0 \, m$.$\Delta T_b = 2 	imes 0.52 	imes 1.0 = 1.04 \, ^{\circ}C$.The boiling point of the solution $T_b = T_b^{\circ} + \Delta T_b$.Since the boiling point of pure water $T_b^{\circ} = 100 \, ^{\circ}C$,we have $T_b = 100 + 1.04 = 101.04 \, ^{\circ}C$.

The molal elevation constant of water is $0.52 \, ^{\circ}C \, kg \, mol^{-1}$. The boiling point of $1.0 \, m$ aqueous $KCl$ solution (assuming complete dissociation of $KCl$) should be $......... \, ^{\circ}C$.

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