Answer:
174.8 g/m is the molar mass of the solute
Explanation:
We must apply colligative property of freezing point depression.
ΔT = Kf . m . i
ΔT = T° freezing pure solvent - T° freezing solution (0° - (-2.34°C) = 2.34°C
Kf = Fussion constant for water, 1.86 °C/m
As ascorbic acid is an organic compound, we assume that is non electrolytic, so i = 1
2.34°C = 1.86°C/m . m
2.34°C / 1.86 m/°C = 1.26 m
This value means the moles of vitamin C, in 1000 g of solvent
We weighed the solute in 250 g of solvent, so let's calculate the moles of vitamin C.
1000 g ___ 1.26 moles
In 250 g ___ (250 . 1.26)/1000 = 0.314 moles
This are the moles of 55 g of ascorbic acid, so the molar mass, will be:
grams / mol ⇒ 55 g/0.314 m = 174.8 g/m
