Question

A solution consists of 3.88 g benzene,C6H6 , and 2.45 g toluene ,C6H5CH3. The vapor pressure of pure benzene at 20. °C is 75 mm Hg and that of toluene at 20.0 °C is 22 mm Hg. Assume that Raoult’s lawholds for each component of the solution, calculate the mole fraction of benzene in the vapor. ( molar mass of benzene= 78.0 g/mole and toluene = 92.0 g/mole.)

Answer= 0.87

Answer 1

moles of benzene = nB = 3.88 / 78 = 0.0497

moles of toulene = nT = 2.45 / 92 = 0.0266

total moles = nB + nT = 0.0763

moles fraction of benzene in liquid phase XB = moles of benzene / total moles = 0.0497/ 0.0763 = 0.65

moles fraction of toulene in liquid phase XT = 0.35

total pressure = XB Po + XT Po

                     = 0.65 x 75 + 0.35 x 22

                     = 56.45 mmHg

P = total = 56.45 mmHg

mole fraction of benzene in vaphor phase = Y_B

X_B P_B = Y_B P

0.65 x 75 = Y_B x 56.45

Y_B = 0.87

mole fraction of benzene in vaphor phase = Y_B = 0.87