A 6.84% (w/v) solution of cane sugar (molar mass = 342 g/mol) is isotonic with a 1.52% (w/v) solution of thiourea. What is the molar mass of thiourea?

Correct answer: 76

Solution

Isotonic solutions: pi1 = pi2 => C1*R*T = C2*R*T => C1 = C2. C (sugar) = (6.84 g/100 mL)/(342 g/mol) = 0.02 mol/100 mL = 0.2 mol/L. C (thiourea) = (1.52 g/100 mL)/M = 0.2 mol/L => M = 1.52/(0.02) = 76 g/mol.

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