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Question
30 g of urea (M = 60 g mol−1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23·8 mm Hg.
Solution
It is given that vapour pressure of water, `p_1^0` = 23.8 mm of Hg
Weight of water taken, w1 = 846 g
Weight of urea taken, w2 = 30 g
Molecular weight of water, M1 = 18 g mol−1
Molecular weight of urea, M2 = 60 g mol−1
Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1.
Now, from Raoult’s law, we have:
`(p_1^0 - p_1)/p_1^0 = n_2/(n_1 - n_2)`
`=> (p_1^0 - P_1)/p_1^0 = (w_2/M_2)/(w_1/M_1 - w_2/M_2)`
`= (23.8 - p_1)/23.8 = (30/60)/(346/18 + 30/60)`
`=> (23.8 - p_1)/23.8 = 0.0105`
=> p1 = 23.5501 mm of Hg
Hence, the vapour pressure of water in the given solution is 23.5501 mm of Hg and its relative lowering is 0.0105.
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