Question

Lowering of vapour pressure due to a solute in 1 molal aqueous solution at 100⁰C is ( assume solute associate not associate)

Answer 1

First of all find mole fraction of gas .
Let mole fraction of gas is x
we know, the relation between mole fraction and molality
molality = 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