Question

Ethanol has a Kb of 1.22 °C/m and usually boils at 78.4 °C. How many mol of an nonionizing solute would need to be added to 48.80 g ethanol in order to raise the boiling point to 85.2 °C.

Answer 1

Answer:

0.272 mol

Explanation:

∆Tb = m × Kb

∆Tb = 85.2°C - 78.4°C = 6.8°C

Kb = 1.22°C/m

mass of ethanol = 48.80 g = 48.80/1000 = 0.0488 kg

Let the moles of non-ionizing solute be y

m (molality) = y/0.0488

6.8 = y/0.0488 × 1.22

y = 6.8×0.0488/1.22 = 0.272 mol