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

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Solution:

$i=2\left( KCl \rightarrow K^{\prime}+ Cl \right)$
$ \Delta T_b=i K_b m=2 \times 0.52 \times 1=1.04$
$\therefore T_b=101.04^{\circ} C$