Calculate the molar mass (in units of g/mol) of the solute.

A solution that contains 55.0 g of ascorbic acid (Vitamin C) in 250. g of water freezes at −2.34°C. Calculate the molar mass (in units of g/mol) of the solute. Kf of water is 1.86°C/m.

M of ascorbic acid (AA) = 55g
M of water = 250g
Normal freezing point of water = 0°C
Lowered freezing point , Fp2 = -2.34°C
Molar mass, Mw of AA = Yg
Kf (water) = 1.86 °C.kg/mol
Solute = AA
Solvent = H2O

ΔTf = i*Kf*m
Where
i = vant hoff factor
m = molality of solute = mol solute / 1000g solvent = mol solute / kg solvent

Nos of moles of AA = (55 / Y)moles
M of H2O = 250g = 250 / 1000 = 0.25 kg

m = [ 55 / Y ] / 0.25 = 55 / 0.25Y
m = 220 / Y

For ascorbic acid, i = 1

ΔT = [ 0 - (-2.34) ]C = 2.34C

Substituting, we have;
2.34 = 1.86*m
2.34 = 1.86m
2.34 / 1.86 = m
1.2581 = m

=> 1.2581 = 220 / Y
Y = 220 / 1.2581
Y = 175g