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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
Given: Vapour pressure of water `(p_1^0)` = 12.3 kPa
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
∴ The mole fraction of the solute in the solution is:
x2 = `1/(1 + 55.56)`
= 0.0177
By using Raoult’s law:
`(p_1^0 - p_1)/p_1^0` = x2
⇒ `(12.3 - p_1)/12.3 = 0.0177`
⇒ 12.3 − p1 = 0.0177 × 12.3
⇒ 12.3 − p1 = 0.2177
⇒ p1 = 12.3 − 0.2177
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.
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