Biphenyl, c12h10, is a nonvolatile, nonionizing solute that is soluble in benzene, c6h6. at 25 °c, the vapor pressure of pure benzene is 100.84 torr. what is the vapor pressure of a solution made from dissolving 13.6 g of biphenyl in 26.4 g of benzene?

One form of Raoult's Law states that the vapor pressure of a solution of a non-volatile solute at certain temperature is equal to the vapor pressure of the pure solvent at the same temperature multiplied by the mole fraction of the solvent, this is:

p = X solvent * P pure solvent,

X solvent = number of moles of solvent / total number of moles.

Here the solute is 13.6 g of C12 H10 and the solvent is 26.4 g C6H6.

=>

moles of solvent = mass in grams / molar mass

molar mass of C6H6 = 6 * 12 g/mol + 6 * 1g/mol = 78 g/mol

moles of solvent = 26.4 g / 78 g/mol = 0.33846 mol

molar mass of C12H10 = 12 * 12g/mol + 10*1g/mol = 154 g/mol

moles of solute = 13.6 g / 154 g/mol = 0.08831 mol

=> X solvent = 0.33846 / (0.33846 + 0.08831) = 0.793

=> p = 0.793 * 100.84 torr = 79.97 torr ≈ 80.0 torr

80.0 torr

Is there a diagram/picture you can provide? it seems this question goes along with a