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JEE Main levelSingle correct optionDifficulty 2 of 3so132

$0.10$ mol of $NaCl$ (assume complete dissociation) is dissolved in $1.00$ kg of water (molar mass $18 g mo l^{- 1}$ ). The relative lowering of vapour pressure is closest to

A $1.80 \times 1 0^{- 3}$
B $5.37 \times 1 0^{- 3}$
C $2.00 \times 1 0^{- 4}$
D $3.59 \times 1 0^{- 3}$

Answer: D

NaCl

gives

0.20

mol of particles. Water

= 1000/18 = 55.56

mol. Relative lowering

= x_{2} = 0.20/ (55.56 + 0.20) = 3.59 \times 1 0^{- 3}

. Ignoring dissociation gives only

1.80 \times 1 0^{- 3}

.

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