A solution has a 1: 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in vapour phase would be:

Question

A solution has a 1: 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in vapour phase would be: (A) 0.786 (B) 0.478 (C) 0.549 (D) 0.200

Tardigrade Admin · Accepted Answer

P=PoC5H12× XC5 H12+PC6 H14o × XC6 H14; p=440×(1/5)+120×(4/5)=(920/5)=184 yc5 H12=(PC5 H12/Pt) Now, YC5 H12=(88/184)=0.478

Q. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at $2 0^{\circ} C$ are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in vapour phase would be:

$0.786$

$0.478$

$0.549$

$0.200$

$P = P_{C_{5} H_{12}}^{o} \times X_{C_{5} H_{12}} + P_{C_{6} H_{14}}^{o} \times X_{C_{6} H_{14}}$ ;
$p = 440 \times \frac{1}{5} + 120 \times \frac{4}{5} = \frac{920}{5} = 184$
$y_{c_{5} H_{12}} = \frac{P _{C_{5} H_{12}}}{P _{t}}$
Now, $Y_{C_{5} H_{12}} = \frac{88}{184} = 0.478$

Q. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20∘C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in vapour phase would be:

0.786

0.478

0.549

0.200