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The $molal$ elevation constant of water $= {0.52}^{\circ} {Ckgmol}^{- 1}$ The boiling point of 1.0 $molal$ aqueous $KCl$ solution (assuming complete dissociation of $KCl)$ , therefore, should be

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${100.52}^{\circ} C$

$98.96 ° C$

${101.04}^{\circ} C$

$99.48 ° C$

answer is B.

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Detailed Solution

$\begin{array}{l} Δ T_{b} = i m k_{b} = 0.52 \times 1 \times 2 = 1.04 \\ T_{b} - T_{b}^{o} = 1.04 \\ ∴ T_{b} = 100 + 1.04 = {101.04}^{\circ} C \end{array}$

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