1.

Claculate the vapour pressure of an aqueous solution of 1.00 molal glucose at 100°C

Answer»

Let mole fraction of gas is x we know, the relation between mole fraction and molalitymolality = x × 1000/(1 - x)M Here, M is Molecular weight of solvent . for aqueous solution , solvent is water . ∴ Molecular weight of water , M = 18g/mol

Now, 1 = x × 1000/(1 - x) × 18⇒18(1 - x) = 1000x ⇒18 = (1000 + 18)x ⇒x = 18/1018

Now, use formula , Relative lowering of vapor pressure = mole fraction of gas ∆P/P₀ = x Here P₀ is initial pressure ,at STP , P₀ = 760 torr so, ∆P = 760 × 18/1018 = 13.44 torr

Hence, lowering of vapor pressure = 13.44 torr


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