Question:
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.
Answer:
Vapour pressure of pure water = 12.3 kPa
We have 1 molal solution that means we have 1 moles of solute in 1kg of solvent.
The formula of vapour pressure when non-volatile liquid is added,
Pressure (P) = vapour pressure of pure liquid P(pure) × molar fraction of liquid (X)
And we have mass = 1 kg and 1 kg is equal to 1000 g
And molar mass of water H2O = 2×1 + 16 = 18 g/mol
Number of moles of water =1000 g / (18g/mol ) =55.56 moles
The formula of molar fraction =55.5/(55.5+1) = 0.9823
Partial fraction = 12.3×0.9823 = 12.08
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