20. The mass of a non-voltaile solute (molarmass 80 g mol-1) which sho

1.	20. The mass of a non-voltaile solute (molarmass 80 g mol-1) which should be dissolvedin 92g of toluene to reduce its vapourpressure to 90%a) 10gb) 20gd) 8.89gc) 9.2 g
Answer» Let the vapour pressure of pure octanebe 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=(w2xM1)/ (M2xw1) ⇒(p10-0.8 p10)/p10=(w2x114)/ (40x 114) ⇒ 0.2p10/p10 =w2/ 40 ⇒ 0.2 =w2/ 40 ⇒w2= 8 g Hence, the required mass of the solute is 8 g.

20. The mass of a non-voltaile solute (molarmass 80 g mol-1) which should be dissolvedin 92g of toluene to reduce its vapourpressure to 90%a) 10gb) 20gd) 8.89gc) 9.2 g

Answer»

Let the vapour pressure of pure octanebe 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=(w2xM1)/ (M2xw1)

⇒(p10-0.8 p10)/p10=(w2x114)/ (40x 114)

⇒ 0.2p10/p10 =w2/ 40

⇒ 0.2 =w2/ 40

⇒w2= 8 g

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