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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.
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Solution
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, the mole fraction of the solute in the solution is
`x_2 = 1/(1 + 55.56) = 0.0177`
Vapour pressure of water, `p_1^0` = 12.3 kPa
Applying the relation, `(p_1^0 - p_1)/p_1^0 = x_2`
`=> (12.3 - p_1)/12.3 = 0.0177`
`=> 12.3 - p_1 = 0.2177`
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.
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