Question

In an experiment, a sample of 4.92 L of nitrogen at 10.8°C and 793 mmHg is bubbled through liquid acetone, C3H6O. The gas plus vapor at its equilibrium partial pressure leaves the liquid at the same temperature and pressure. If 5.260 g of acetone has evaporated, what is the vapor pressure of acetone at 10.8°C?

Answer 1

At 10.8 deg C , the pure vapor pressure of acetone = 123 mmHg

Molefraction of acetone = Number of mols of acetone / Total mols

From the given data , moles of N2 = PV/RT = (793 /760) atm * 4.82 L / 0.0821 *284.3 = 0.215 mol

Similaryl, number of mols of acetone = 5.260 g/58 = 0.090 mol

Thus, total mols = 0.215 mol + 0.090 mol = 0.305 mol

mole fraction of acetone = 0.215 mol /0.305 mol = 0.704

mole fraction of N2 = 0.088 / 0.305 = 0.288

Thus Vapor pressure of acetone = (Vapor pressure at Pure state)(mole fraction)

                                                   = (123 mmHg)(0.704) = 86.7 mmHg