Question

A solution that contains 55.0 g of ascorbic acid (vitamin C) in 250.0 g of water freezes at –2.34°C.

Answer 1

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