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NEET levelSingle correct optionDifficulty 1 of 3so040

$18 g$ of glucose (molar mass $180$ ) is dissolved in $90 g$ of water (molar mass $18$ ). The relative lowering of vapour pressure of water is

A0.0200
B0.0196
C0.1
D0.9804

Answer: B

n_{2} = 18/180 = 0.10

and

n_{1} = 90/18 = 5.0

.

(p^{0} - p) / p^{0} = n_{2} / (n_{1} + n_{2}) = 0.10/5.10 = 0.0196

. Using

n_{2} / n_{1} = 0.0200

is only an approximation.

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