Vapour pressure of C6H6 and C7H8 mixture at 50 degree C is given P(mm Hg)= 180Xb+90, where Xb is the mole fraction of C6H6. A solution is prepared by mixing 936 gm benzene and 736 gm toluene and if vapours over this solution are removed and condensed into liquid and again brought to temperature 50 degree C, what would be mole fraction of C6H6 in the vapour state.

29-Apr-2016 1:01 PM
~4

0

Given: mass of C6H6  (wb)= 936 g
​​            mass of C7H8 (wa) = 736 g
​            pressure of mixture, p = 180 X+ 90 mmHg
To calculate: mole fraction of benzene in vapour state

Solve: Mole fraction of benzene in liquid state (Xb) is given by
$X_{b} = \frac{n_{b}}{n_{b} + n_{a}}$
where nand nb are number of moles of toluene and benzene respectively.
Number of moles is given by
$n = \frac{w}{M}$
where w is the given mass and M is molar mass of the substance.

Molar mass of Benzene (Mb) = 78 g mol-1
Molar mass of Toluene (Ma) = 92 g mol-1

nb = wb/Mb

$n_{b} = \frac{936 g}{78 g mol^{-1}}$

nb = 12 mol

na = wa/Ma

$n_{a} = \frac{936 g}{92 g mol^{-1}}$

na = 8 mol

$X_{b} = \frac{n_{b}}{n_{b} + n_{a}}$

$X_{b} = \frac{12}{8 + 12}$

Xb = 0.6

and Xa = 1 - Xb
​              = 1 - 0.6
=0.4

Thus substituting the value of Xb in the given equation of pressure

p = 180 Xb  + 90 mmHg

p = 180 × 0.6 +90 mmHg

p = 198 ​mmHg

Comparing the given equation of pressure
p = (180 Xb  + 90 ) mmHg
with the standard equation of

$p = p_{a}^{0} + (p_{b}^{0} - p_{a}^{0} )X_{b}$

$p_{a}^{0} = 90 mmHg$

$(p_{b}^{0} - p_{a}^{0}) = 180 mmHg$

thus, $p_{b}^{0} = 90 mmHg$

mole fraction of benzene in the vapour state is given by
$y_{b} = p_{b}/P$and

$p_{b} = p_{b}^{0} X_{b}$

where $p_{b}^{0}$ is the partial pressure of pure benzene at 50 deg. C.
pb= 90 ×0.6
= 54 mmHg
Thus
yb= 54/198    = 0.27
Therefore, the mole fraction of benzene in vapour state is 0.27.

29-04-2016 23:14
~32

0

( PS I had edited your question because the image you had added was too large. Please type the questions in such cases)

29-04-2016 23:15
~32

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