Sol. Using the formula
PoPo−Ps=MolesofsolventMolesofsolute(for dilute solution)
Po=Vapour pressure of pure water
Ps=Vapour pressure of solution
54.254.2−54=18100w/180w=54.20.2×1000=54.2200=3.69
JEE Main 2023 — Chemistry Physical Chemistry
What weight of glucose must be dissolved in 100g of water to lower the vapour pressure by 0.20mmHg?
(Assume dilute solution is being formed)
Given: Vapour pressure of pure water is 54.2mmHg at room temperature.Molar mass of glucose is 180gmol−1
Held on 11 Apr 2023 · Verified 6 Jul 2026.
3.59g
3.69g
4.69g
2.59g
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