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Boiling point of chloroform is 61°C. After addition of 5.0 g of a non-volatile solute to 20 g chloroform boils at 64.63°C. If kb = 3.63 K kg mol^-1, what is the molecular weight of the solute?(a) 320 g/mol(b) 100 g/mol(c) 400 g/mol(d) 250 g/molThis question was posed to me during an internship interview.My question is from Colligative Properties and Determination of Molar Mass in division Solutions of Chemistry – Class 12 |
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Answer» RIGHT answer is (d) 250 g/mol To explain: Given, b.p. of chloroform = 61°C New b.p. after ADDITION = 64.63°C Mass of solute, w2 = 5.0 g Mass of solvent, W1 = 20 g Kb = 3.63 K kg mol^-1 From these, ∆Tb = 64.63 – 61 = 3.63°C Using ΔTb = (kb x 1000 x w2)/(M2 x w1) M2 = (kb x 1000 x w2)/(ΔTb x w1) M2 = (3.63 x 1000 x 5)/(3.63 x 20) = 250 g/mol. |
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