At a given temperature, you have a mixture of benzene (vapor pressure of pure be

Question

At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of pure toluene = 290 torr). The vapor is collected, condensed and the resulting liquid is allowed to equilibrate with its vapor. This vapor is collected, condensed and the resulting liquid is allowed to equilibrate with its vapor. The mole fraction of benzene in this vapor is 0.830. Assuming ideal behavior, calculate the mole fraction of toluene in the original solution.

Explanation / Answer

Let Xa and Xb be mole fractions of benzene and toulene in its solution

and Ya and Yb be mole fractions of benzene and toulene in its vapour.

pao and pbo are vapour pressures of pure benzene and pure toulene

paand pb are partial pressures of vapours of benzene and toulene

and P is total pressure.

Then Ya = pa / P   = Xa pao / P      ------------- (1)

Yb = pb / P = Xb pbo / P   ---------------(2)

Devide eq (1) by (2)

Ya / Yb = Xa pao / Xb pbo

Given Ya = 0.830, then Yb = 1 - 0.830 = 0.17 ,

pao = 745 torr , pbo = 290 torr and Xa + Xb = 1

So 0.830 / 0.17 = (1 - Xb)745 / Xb 290

       4.8823 * 290 * Xb = 745 - 745 Xb

2160.867 Xb = 745

Xb = 745 / 2160.867

      = 0.344769

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