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Concentration Terms - An Introduction (w/w etc)

A solution ha...

Question

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:

0.786

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0.478

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0.549

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0.200

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Solution

The correct option is C $0.478$
mole fraction in vapor phase
$Y_{A} = \frac{x_{A} p_{A}}{p_{B} + (p_{A} - p_{B}) x_{A}}$

Given that

$n_{A} = 1, n_{B} = 4$

$n_{A} = \frac{1}{1 + 4} = \frac{1}{5} = 0.2$

$n_{B} = 1 - 0.2 = 0.8$

$p_{A} = 440, p_{B} = 120$

$y_{A} = \frac{0.2 \times 440}{120 + (440 - 120) \times 0.2}$

$= \frac{88}{120 + 64} = \frac{88}{184}$

$y_{A} = 0.478$

Hence, the correct option is $B$

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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) 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?

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

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

The correct option is C 0.478mole fraction in vapor phase YA=xApApB+(pA−pB)xAGiven thatnA=1,nB=4nA=11+4=15=0.2nB=1−0.2=0.8pA=440,pB=120yA=0.2×440120+(440−120)×0.2=88120+64=88184yA=0.478Hence, the correct option is B

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: