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?

simple question:

 

Here solute is Urea and solvent is water.

Thus

Relative lowering is the quantity on the RHS in the first equation

How u got p1=27.81

See dude, is the change in pressure, so the new pressure is given by .

(We subtract as the vapour pressure is lowered)

Thank u dost

mole fraction of water   =  .983

so relative lowering = mole fraction of urea = 1 - .983 = .017

and V.P of water 23.8 * .983 = 23.39 mm Hg