Question

13. A mixture of benzene and toluene contains
30% by mass of toluene. At 30°C, vapour
pressure of pure toluene is 36.7 mm Hg
and that of pure benzene is 118.2 mm Hg.
Assuming that the two liquids form ideal
solutions, calculate the total pressure and
partial pressure of each constituent above
the solution at 30°C. (86.7 mm, P = 96.5
3.​

Answer 1

Answer:

YOUR QUESTION :

A mixture of benzene and toluene contains

30% by mass of toluene. At 30°C, vapour

pressure of pure toluene is 36.7 mm Hg

and that of pure benzene is 118.2 mm Hg.

Assuming that the two liquids form ideal

solutions, calculate the total pressure and

partial pressure of each constituent above

the solution at 30°C. (86.7 mm, P = 96.5).

SOLUTION :

Given:

Mass % of toluene = 30%

Vapour pressure of pure toluene, Pt° = 36.7 mm Hg

Vapour pressure of pure benzene, Pb° = 118.2 mm Hg

To Find:

The total pressure and partial pressure of each constituent.

Calculation:

Let the total mass be x

=> Mass of toluene = 0.3 x

=> Mass of benzene = 0.7 x

- Mole fraction of toluene Xt = (0.3x/92) / {(0.3x/92) + (0.7x/78)}

=> Xt = 0.266

- Mole fraction of benzene Xb = 1 - 0.266 = 0.734

- Partial pressure of toluene, Pt = Pt° × Xt

=> Pt = 36.7 × 0.266

=> Pt = 9.76 mm Hg

- Partial pressure of benzene, Pb = Pb° × Xb

=> Pb = 118.2 × 0.734

=> Pb = 86.76 mm Hg

= Total Pressure P = Pt + Pb

=> P = 9.76 + 86.76

⇒ P = 96.52 mm Hg

So the partial pressure of toluene is 9.76 mm Hg, the partial pressure of benzene is 86.76 mm Hg and the total pressure is 96.52 mm Hg.

Answer 2

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