Question

Vapour pressure of water is 360 mm Hg , how much urea should be added to 200 ml water to reduce its vapour pressure by 0.5%?

Answer 1

Vapor Pressure of pure water = p0=360mm Hg 

Lowered Vapor Pressure, p= 360-0.5% of 360 

= 360-1.8 

= 358.2 mm Hg 

Weight of water w1= 200mlx1gml=200g 

Weight of Urea w2= ? 

Molecular weight of water, M1= 18g/mol 

Molecular weight of Urea, M2= 60g/mol 

Accoding Roult's Law: 

P0-p/p = n2/n1+n2 

360-358.2/358.2= w2/M2 divided by w1/m1 + w2/m2 

0.005=w2/m2 divided by w1/m1+w2/m2 

0.005/0.995 x w1/m1 x M2 = w2 

0.005 x 200/18 x 60 = w2 

W2 = 3.33g 

3.33g Urea should be added.

Answer 2

Answer

According to Raoult's Law, the relative lowering of vapour Pressure is equal to the mole-fraction of the Solute.

So( P°-P)/P°= x2 =n2/(n1+n2),

since n1>>> n2

n1+n2~= n1

So ( P°-P)/ P° =(w2/m2)/(w1/m1)

Now P°-P= 0.5, P°= 100.

So 0.5/100 = (w2/60)/(200/18).

So 0.005= (w2/60/(200/18).

Upon solving we get w2= 3.33 g of Urea

Hope it helps..