A solution was made by dissolving 3.75 g of a pure non volatile solute in 95 g of acetone. The boiling point of pure acetone was observed to be 55.95°c, and that of the solution was 56.50°c. If the molar boiling point elevation constant of acetone is 1.71°c kg/mol, what is the approximate molar mass of the solute?

2 answers

delta T = i*Kb*m
delta T = 56.50 - 55.95 = ?
i = 1
Kb = 1.71
m = molality. Plug delta T, i and Kb as I've listed them and solve for molality.
Then molality = m = mol/kg solvent. Kg solvent = 95 g acetone. Use 0.095 kg. You have kg and you have molality. Solve for mol. Then mols = grams/molar mass. You have mol and grams, solve for molar mass. Post your work if you get stuck.
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
58
95

=1.64moled

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

×
1.71
328.95×329.5

=3676.35