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JEE Main levelSingle correct optionDifficulty 2 of 3so216

$31 g$ of ethylene glycol (molar mass $62 g mo l^{- 1}$ , non-volatile) is dissolved in $500 g$ of water. With $K_{b} = 0.52 K kg mo l^{- 1}$ for water and a normal boiling point of $10 0^{\circ} C$ , the boiling point of the solution is

A100.26 $^{\circ} C$
B100.52 $^{\circ} C$
C101.04 $^{\circ} C$
D99.48 $^{\circ} C$

Answer: B

m = (31/62) /0.500 = 1.00 mol k g^{- 1}

.

Δ T_{b} = 0.52 \times 1.00 = 0.52 K

, and the boiling point rises, so

T_{b} = 100 + 0.52 = 100.5 2^{\circ} C

.

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