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Question
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
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Solution
Given: P0 = 23.8 mm Hg
w1 = 850 g, M1 (water) = 18 g mol−1
w2 = 50 g, M2 (urea) = 60 g mol−1
Ps = ?,
`(P^0 - P_s)/P^0` = ?
∴ n1 = `850/18` = 47.22
∴ n2 = `50/60` = 0.83
Applying Raoult’s law:
`(P^0 - P_s)/P^0 = n_2/(n_1 + n_2)`
or, `(P^0 - P_s)/P^0 = 0.83/(47.22 + 0.83)`
`(P^0 - P_s)/P^0 = 0.83/48.05`
`(P^0 - P_s)/P^0` = 0.017
Thus, the relative lowering of vapour pressure = 0.017
Again, `(Delta P)/P^0` = 0.017
∴ ΔP = 0.017 × 23.8
or, P0 − Ps = 0.017 × 23.8
or, Ps = 23.8 − (0.017 × 23.8)
or, Ps = 23.4 mm Hg
Thus, the vapour pressure of water in the solution is 23.4 mm Hg.
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