Question

A mixture of benzene and toluene contains 30% by mass of toluene at 30 degree celcius by vapour pressure of pure toulene 36.7 mm of hg and that of pure benzene is 118.2 mm of hg.assuming that two liquids form ideal solutions ,calculate the total pressure and partial pressure of each constituent above the solution at 30degree celcius

Answer 1

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

Answer:

So, these is the answer

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