In this case, according to the Raoult's law, we can find the mole fraction of 79 wt% benzene and 21 wt% toluene at 50 °C in the vapor phase as follows:
[tex]y_iP=x_iP_i[/tex]
However, we first have to calculate the mole fractions in the solution as follows (b stands for benzene and t for toluene):
[tex]\\x_b=0.79\frac{g\ benzene}{g\ solution} *\frac{(78.11*x_b+92.14*x_t)}{1mol\ solution} *\frac{1mol\ toluene}{78.11g\ toluene} \\\\x_b=0.79\frac{g\ benzene}{g\ solution} *\frac{78.11*x_b+92.14*(1-x_b)}{1mol\ solution} *\frac{1mol\ toluene}{78.11g\ toluene}\\\\x_b=0.79*(x_b+1.18*(1-x_b))\\\\x_b=0.79x_b+0.932-0.932x_b\\\\x_b=\frac{0.932}{1+0.932-0.79} =0.816\\\\x_t=1-x_b=1-0.816=0.184[/tex]
Next, we calculate the total pressure as follows, according to the Dalton's law:
[tex]P=x_bP_b+x_tP_t=0.816*271mmHg+0.184*91.5mmHg=237.972mmHg[/tex]
Finally, the mole fractions of the vapor phase turn out:
[tex]y_b=\frac{0.816*271mmHg}{237.972mmHg}=0.929\\\\y_t=1- 0.929=0.071[/tex]
Learn more:
