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Vapour pressure of pure water at …

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Vapour pressure of pure water at 298k is 23.8mm hg. 50gm of urea (NH2CONH2) is dissolved in 850g of water.calculate the vapour pressure of water for this solution and it's relative lowering.

Posted by Amisha Rajahansa 2 years, 4 months ago

CBSE > Class 12 > Chemistry

1 answers

Ritumoni Borgohain 2 years, 4 months ago

Vapour pressure of water,P1=23.8 m of hg The weight of water=850 g The weight of Urea=50 g The molecular weight of water(H_2O)=1×2+16=18 g mol−1 molecular weight of urea (NH2CONH2)=2N+4H+C+O=2×14+4×1+12+16=60 g mol−1 Use formula number of moles=molar massmass number of moles of water n1=18850=47.22 number of moles of urea n2=6050=0.83 Now, we have to calculate the vapour pressure of water in the solution. we take vapour pressure as P1. Use the formula of Raoult's law P10P10−P1=n1+n2n2 plug the values we get 23.823.8−P1=47.22+0.830.83 23.823.8−P1=0.0173 23.8−P1=23.8×0.0173 P1=23.4 m Hg Vapour Pressure of water in the given solution=23.4 mm of Hg Relative lowering=0.0173

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