Question

25.0 g of a molecular compound is dissolved in 50.0 g of water giving the resulting solution a boiling point of 104 ˚C. Calculate the molar mass of the solution. K
b
of water is 0.52 ˚C/m.
Multiple Choice
65.0 g/mol
75.4 g/mol
107.6 g/mol
122.3 g/mol

Answer 1

ΔT = T(boiling solution) – T(water boiling)

ΔT = 104 – 100 = 4 ºC

$m = \frac{ΔT}{K_b} \\ m = \frac{4}{0.52} = 7.69 \;mol/kg \\ m = \frac{moles\;solute}{kg\;solvetnt}$

kg solvent = 0.05 kg

moles solvent $= m \times kg\;solvent$

moles solvent $= 7.69 \times 0.05 = 0.3845 \;mol$

$m_c = 25.0 \;g \\ n_c = 0.3845 \;mol \\ MM_c = \frac{25.0}{0.3845} = 65.02 \;g/mol$

Answer: 65.02 g/mol