Question

What weight of solute having molecular weight 60 is required?

Answer 1

Answer:Let the vapour pressure of pure octane be p10.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is 80/100 p10 = 0.8 p10.

Molar mass of solute, M2 = 40 g mol - 1

Mass of octane, w1 = 114 g

Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1 = 114 g mol - 1

Applying the relation,

(p10 - p1) / p10 = (w2 x M1 ) / (M2 x w1 )

⇒ (p10 - 0.8 p10) / p10 = (w2 x 114 ) / (40 x 114 )

⇒ 0.2 p10 / p10 = w2 / 40

⇒ 0.2 = w2 / 40

⇒ w2 = 8 g

Explanation: