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.​

Answer 1

vapour pressure of water ( pi ) = 23.8 mm Hg

weight of water ( W ) = 850 g

weight of urea ( w ) = 50 g

molar mass of water ( M ) = 18 g / mol

molar mass of urea ( m ) = 60 g / mol

so ,

mole of water ,

N = W / M = 850 / 18 = 47.22 mol

mole of urea ,

n = w / m = 50 / 60 = 0.83 mol

Let vapour pressure of water in solution , using RAULT'S theorem :

pi - pI / pi = n / n + N

(23.8 - pI) / 23. 8 = 0.83 / ( 0.83 + 47.22)

( 23.8 - pI) / 23.8 = 0.83 / 48.05

( 23.8 - pI) / 23.8 = 0.0173

pI = 23.44 mm Hg

Answer 2

Answer:

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