Question

What weight of non volatile solute (M.wt. 60 ) is required to dissolve in 180g 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

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Answer 2

Given:

M. wt. = 60

W = 180 gm

Vapour pressure, P = 4/5 of pure water = 4/5 P°

To Find:

The weight of non- volatile solute (w).

Calculation:

⇒ No of moles of solute, n1 = w/60

- No of moles of water, n2 = 180/18 = 10

- We Know that:

Relative lowering of V. P. = Mole fraction of solute

⇒ (P° - P) / P° = X

⇒ (P° - 4/5 P°) / P° =  (w/60) / { (w/60) + 10}

⇒ 1/5 =  w / (w + 600)

⇒ w + 600 = 5w

⇒ 4w = 600

⇒ w = 150 gm

- So, the weight of non- volatile solute is 150 gm.