Question

A solution consists of a mixture of benzene and toluene and is allowed to come to equilibrium with its vapor. The vapor is then condensed and found to contain equal moles of each component. Calculate the composition (mole percent) of benzene in the original solution. The vapor pressures of pure benzene and toluene at this temperature are 750. torr and 300. torr respectively.

Answer 1

This problem is describing the vapor-liquid equilibrium between benzene and toluene in which equal moles of each component are present in the vapor and the initial compositions in the liquid are required, considering that the vapor pressures of benzene and toluene are 750. torr and 300. torr respectively.

In this case, it turns out possible to use the Raoult's law in order to come up with a model for this system:

$y_BP_t=x_BP_B\\\\y_TP_t=x_TP_T$

Thus, we infer that the compositions of benzene and toluene in the vapor are both 0.5 because equal number of moles are condensed. Next, since the mole fractions in the liquid phase sum 1, we rewrite the previous expressions as follows:

$y_BP_t=x_BP_B\\\\y_TP_t=(1-x_B)P_T$

And insert one into the other, via the total pressure:

$y_B[\frac{(1-x_B)P_T}{y_T} ]=x_BP_B$

Now, we can start plugging in the numbers we were given:

$0.5[\frac{(1-x_B)300.torr}{0.5} ]=x_B*750.torr$

Then, solve for the mole fraction of benzene in the liquid phase:

$=x_B*750.torr\\\\300.torr-300.torr*x_B=750.torr*x_B\\\\x_B=\frac{300torr}{1050torr} =0.286$

And finally, the mole fraction of toluene in the liquid phase:

$x_T=1-0.286=0.714$

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