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JEE Main levelSingle correct optionDifficulty 2 of 3so044

$0.10 mol$ of $NaCl$ is dissolved in $90 g$ of water (molar mass $18$ ). Assume complete dissociation and treat every ion as an independent solute particle. The relative lowering of vapour pressure is

A0.0196
B0.0385
C0.0400
D0.0192

Answer: B

Solute particles

= 2 \times 0.10 = 0.20 mol

; water

= 5.00 mol

.

(p^{0} - p) / p^{0} = 0.20/ (5.00 + 0.20) = 0.0385

. Counting only

0.10

particles gives 0.0196.

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