Answer to Question #310756 in General Chemistry for Janelle S

Question

A solution was produced by dissolving 3.75 g of a nonvolatile solute in 95g of acetone. The boiling point of pure acetone was observed to be 55.95 ºC, while the boiling point of the solution was 56.50ºC. If the boiling point for acetone is 1.71ºC/m, what is the approximate molar mass of solute?

Step 1: First compute the molality of the boiling point equation.

Step 2. Then, from the definition of molality, compute the number of moles solute, n(solute), in the sample.

Step 3. Solve for the molar mass. 

Answer 1

∆T = K_b × m

∆T = T_sol - T_pure = 56.50°C - 55.95°C = 0.55°C

m = n(solute) / kg (solvent) = m(solute)/ (Mr(solute)×kg(solvent)) = 3.75 g / (Mr(solute)×0.095 kg)

0.55°C = 1.71ºC/m × 3.75 g / (Mr(solute)×0.095 kg)

Mr(solute) = 122.7 g/mol