The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molar solution of a non-volatile solute in it.

Solution

Answer:

1 molar solution means 1 number of moles of the solute per kilogram (kg) of the solvent. Mole fraction of solute.

= \frac{1}{1 + \frac{1000}{18}} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \frac{P_{A}^{0} - P_{A}}{P_{A}^{0}} = x_{B}

\frac{12.3 - P_{A}}{12.3} = \frac{1}{56.55}

12.3 - P_{A} = \frac{12.3}{56.55} P_{A} = 12.3 - \frac{12.3}{56.55} = 12.3 - 0.2175 P_{A} = 12.0825 kpa = 12.08 kpa

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