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JEE Main levelSingle correct optionDifficulty 2 of 3so192

A $0.200 mol k g^{- 1}$ aqueous solution of $CaCl X_{2}$ is $80%$ dissociated. With $K_{b} = 0.52 K kg mo l^{- 1}$ and a normal boiling point of $10 0^{\circ} C$ for water, the boiling point of the solution is closest to

A100.27 $^{\circ} C$
B100.19 $^{\circ} C$
C100.31 $^{\circ} C$
D100.10 $^{\circ} C$

Answer: A

CaCl X_{2} Ca X^{2 +} + 2 Cl X^{-}

so

i = 1 + 2 α = 1 + 2 (0.80) = 2.60

.

Δ T_{b} = i K_{b} m = 2.60 \times 0.52 \times 0.200 = 0.2704 K

. The boiling point is

100 + 0.2704 = 100.2 7^{\circ} C

.

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