Question

what weight of solute of molecular weight 60 is required to dissolve in 180gm of water to reduce the vapour pressure to 4/5th of pure water

Answer 1

Let the weight of solute (w2) = w grams

Weight of solvent (w1)= 180 grams

Molar mass of solvent H2O(M1) = 18

Moalr mass of solute (M2) = M

Let the vapour pressure of pure solvent (p1°) = x

Vapour pressure of solution (p1) = (x*4)/5

(p1°- p1)/ p1°= (w2*M1)/(w1*M2)

(x- 4x/5)x = (w *18)/(180*M2)

(1-4/5) = w/(10*M)

w=1/5*10*M

w= 2M

Therefore weight of the solute = 2M