The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol - 1
∴ Number of moles present in 1000 g of water = 1000/18
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
x2 = 1 / (1+55.56) = 0.0177.
It is given that,
Vapour pressure of water, p10 = 12.3 kPa
Applying the relation, (P10 - P1) / P10 = X2
⇒ (12.3 - p1) / 12.3 =0.0177
⇒ 12.3 - P1 = 0.2177
⇒ p1 = 12.0823
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.
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