Answer to Question #124311 in Physical Chemistry for giang

Question

The vapour pressure of a sample of benzene is 53.0 kPa
at 60.6°C, but it fell to 51.2 kPa when 0.133 g of an organic
compound was dissolved in 5.00 g of the solvent. Calculate
the molar mass of the compound.

Answer 1

Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present:

P_solution = χ _solvent P^o_solvent

χ _solvent = P_solution / P^o_solvent = 51.2 / 53.0 = 0.966

χ _solvent = moles benzene / total moles

M (C₆H₆) = 78.11 g / mol

n = 5.00 / 78.11 = 0.064 mol

M (A) = x g / mol (unknown)

n = 0.133 / x

Put all data together:

$$0.966 = 0.064 / (0.064 + 0.133/x)$$

From the equation:

x = 58.7 (g/mol) - molar mass of A