Question

The mass of a non volatile solute of molar mass 40 g per mole that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is??

Answer 1

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

Hence, the required mass of the solute is 8 g.

Answer 2

answer in the attachment
And yes the answer is 10g...
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