Answer:
Molar mass for the unknown solute is 109 g/mol
Explanation:
Freezing point depression is the colligative property that must be applied to solve the question.
T°F pure solvent - T°F solution = Kf . m
Let's analyse the data given
Camphor → solvent
Unknown solute → The mass we used is 0.186g
T°F pure solvent = 179.8°C and T°F solution = 176.7°C.
These data help us to determine the ΔT → 179.8°C - 176.7°C = 3.1°C
So we can replace → 3.1°C = 40°C/m . m
m = 3.1°C / 40 m/°C → 0.0775 mol/kg
We have these moles of solute in 1kg of solvent, but our mass of camphor is 22.01 g (0.02201 kg).
We can determine the moles of solute → molality . kg
0.0775 mol/kg . 0.02201 kg = 1.70×10⁻³ moles
Molar mass → mass (g) / moles → 0.186 g / 1.70×10⁻³ mol = 109 g/mol
