Answer:
Molar mass of solute is 89.28 g/m
Explanation:
Colligative property of freezing point depression to solve this:
ΔT = Kf . m . i
i = number of particles, dissolved in solution. In this case, it is a nonelectrolyte, so i = 1.
m = molalilty (mol of solute/1kg of solvent
ΔT = T° freeze pure solvent - T° freeze solution
0°C - (-2.50°C) = 1.86 °C/m . m
2.50°C / 1.86 m/°C = m
1.34 mol solute/kg solvent = m
This means, that in 1000 g of solvent, we have 1.34 moles but we have 250 g of solvent, so let's make a rule of three.
1000 g ____ 1.34 moles
250 g _____(2.50 . 1.34) / 1000 = 0.336 moles
To find the molar mass, we divide mass / moles
30 g/ 0.336 moles = 89.28 g/m
