The vapour pressure of acetone at 20^oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20^oC, its vapour pressure was 183 torr. What is the molar mass (g/mol) of the substance?

64

\(\frac{Po - P}{Po}\) = X_b = \(\frac{Nb}{Na + Nb}\) ≈ \(\frac{Nb}{Na}\) ≈ \(\frac{Wb X Ma}{Wa X Mb}\)

\(\frac{185-183}{185}\) = \(\frac{1.2 X 58}{100 X Mb}\)

M_b = \(\frac{1.2 X 58 X 185}{2 X 100}\) = 64.38 ≈ 64