Engineering Board Exam Reviewer · Chemistry · hard
Dissolving 2 g of a nonelectrolyte in 100 g of water lowers the freezing point by 0.372 °C. What is the molar mass of the solute? (Kf = 1.86 ° C/m)
- 10 g/mol
- 100 g/mol
- 50 g/mol
- 200 g/mol
Answer
B. 100 g/mol
Solution
m = Tf/Kf = 0.372/1.86 = 0.20 mol/kg. n = 0.20(0.1 kg) = 0.020 mol. M = 2/0.020 = 100 g/mol.
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