JEE Main levelSingle correct optionDifficulty 2 of 3so178
The vapour pressure of pure water at 25∘C is 23.8mmHg. A solution is prepared by dissolving 90g of glucose (molar mass 180gmol−1) in 180g of water (molar mass 18gmol−1). Taking glucose as non-volatile, the vapour pressure of the solution in mmHg is closest to
A24.93
B22.61
C22.67
D1.13
Answer: C
Raoult's law: p=x1p0. Moles of glucose =0.50, moles of water =10.0, so x1=10/10.5=0.9524. Then p=23.8×0.9524=22.67mmHg.