Answer to Question #185635 in General Chemistry for Joshua

Question

A solution was produced by dissolving 3.75 g of a nonvolatile solute in 95g of acetone. The boiling of pure acetone was observed to be 55.95 °C, while the boiling point of the solution was 56.50°C. If the boiling point for acetone is 1.71°C/mz 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

Given: mass of solute = 3.75 g

mass of solvent = 95 g = 0.095 kg

boiling point of solvent = 55.95°C

boiling point of solution = 56.50°C

Kb = 1.71°C·kg/mol .

Moles of acetone = $\frac{95}{58}=1.64 moled$

Molarity $= \frac{58}{1000}×\frac{328.95×329.5}{1.71}= 3676.35$