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.
It is given that vapour pressure of water, `p_1^0`= 23.8 mm of Hg
Weight of water taken, w1 = 850 g
Weight of urea taken, w2 = 50 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` = `(50/60)/(850/18 + 50/60`
`=>(23.8-p_1)/23.8` = `0.83/(47.22+0.83)`
`=> p_1 = 23.4 "mm of Hg"`
Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173