A solution has a 1-4 mole ratio of pentane to hexane- The vapour pressure of the pure hydrocarbons at 20-C are 440 mm of Hg for pentane and 120 mm of Hg for hexane- The mole fraction of pentane in the vapour phase would be-

The correct option is D 0.478 Mole fraction of pentane solution= 11+4=15 Mole fraction of hexane solution= 41+4=45 Total pressure of solution, ...

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Standard XII

Chemistry

Liquids in Liquids

A solution ha...

Question

A solution has a $1 : 4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20^{\circ} C$ are $440$ mm of $H g$ for pentane and $120$ mm of $H g$ for hexane. The mole fraction of pentane in the vapour phase would be?

0.549

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0.200

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0.786

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0.478

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Solution

The correct option is D $0.478$
Mole fraction of pentane solution= $\frac{1}{1 + 4} = \frac{1}{5}$

Mole fraction of hexane solution= $\frac{4}{1 + 4} = \frac{4}{5}$

Total pressure of solution, $P_{S} = X_{P} P_{P}^{O} + X_{H} P_{H}^{O}$

$= \frac{1}{5} \times 440 m m H g + \frac{4}{5} \times 120 m m H g$
$= 184 m m H g$

$∴$ The fraction of pentane in the vapour phase

$= \frac{X_{P} P_{P}^{O}}{P_{S}} = \frac{88}{184} = 0.478$ .

Hence, the correct option is $D$

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Similar questions

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°Care 440mmHg for pentane and 120mm Hg for hexane. The mole fraction of pentane in the vapour phase would be:

Q. A solution has a

1 : 4

mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at

20^{o}

C are

440

mm Hg for pentane and

120

mm Hg for hexane. The mole fraction of pentane in vapour phase would be:

Q. Calculate vapour pressure of a mixture containing

252 g

n -

pentane (MW=72) and

1400 g

n -

heptane (MW=100) at

20^{o} C

. The vapour pressure of

n -

pentane

= 420

mm of

H g

and

n -

heptane

= 36

mm of

H g

Q. (a) The vapour pressure of n-hexane and n-heptane at

273 K

are

45.5 m m H g

and

11.4 m m H g

, respectively.What is the composition of a solution of these two liquids if its vapour pressure at

273 K

37.3 m m H g

.
(b) The mole fraction of n-hexane in the vapour above a solution of n-hexane and n-heptane is

0.75

273 K

.
What is the composition of the liquid solution?

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

The correct option is D 0.478Mole fraction of pentane solution= 11+4=15Mole fraction of hexane solution= 41+4=45Total pressure of solution, PS=XPPOP+XHPOH =15×440 mm Hg+45×120 mm Hg =184 mm Hg∴ The fraction of pentane in the vapour phase =XPPOPPS=88184=0.478 .Hence, the correct option is D

A solution has a $1 : 4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20^{\circ} C$ are $440$ mm of $H g$ for pentane and $120$ mm of $H g$ for hexane. The mole fraction of pentane in the vapour phase would be?