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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. - CBSE (Science) Class 12 - Chemistry

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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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Solution 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. Concept: Vapour Pressure of Liquid Solutions - Vapour Pressure of Solutions of Solids in Liquids.
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