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One gram of a solute dimerizes 75% in 100 g of water, and the observed depression in freezing point is 0.093 degree C. Find the average molar mass of the solute after dimerisation (in g/mol). [Kf(H2O) = 1.86 K*kg/mol]

1. 40
2. 80
3. 200
4. 100

Correct answer: 200

Solution

Let M = molar mass of monomer. Original moles in 1g = 1/M. Let 75% dimerize: moles that dimerize = 0.75/M; they form 0.375/M dimer molecules. Remaining monomer = 0.25/M. Total moles after dimerization = 0.25/M + 0.375/M = 0.625/M. Molality = (0.625/M) / 0.1 kg = 6.25/(M) mol/kg. Delta_Tf = Kf * m: 0.093 = 1.86 * 6.25/M. M = 1.86*6.25/0.093 = 11.625/0.093 = 125 g/mol. Average molar mass after dimerization = total mass / total moles = 1g / (0.625/M) = M/0.625 = 125/0.625 = 200 g/mol.

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