When 20.0 grams of an unknown compound are dissolved in 500. grams of benzene, the freezing point of the resulting solution is 3.77^$\circ$C. The freezing point of pure benzene is 5.444^$\circ$C, and the K_f for benzene is 5.12^$\circ$C/m. What is the molar mass of the unknown compound A) 110 g/mol B) 120 g/mol C) 130 g/mol D) 140 g/mol E) None of the above

Question

Quizplus · Accepted Answer

ΔTf = Kf x m
m = molality of the solution = moles of solute/kg of solvent

First convert the freezing point measurements to Kelvin:
ΔTf = 5.444°C - 3.77°C = 1.674°C = 1.674 K
Kf = 5.12°C/m = 5.12 K/m

m = (20.0 g/molar mass) / (0.5 kg) = (0.04 molar mass) / (kg solvent)

Now substitute into the equation:
1.674 = 5.12 x (0.04/molar mass)
molar mass = 120 g/mol

When 200 Grams of an Unknown Compound Are Dissolved in 500