Question

A pure solvent freezes at 6.35 degrees Celsius and a pure solute freezes at 832.0 degrees Celsius. The molar mass of the solvent is 96.4 g/mol and the molar mass of the solute is 78.5 g/mol. The freezing point depression constant for the solvent is 6.33 degrees/m. A solution is made of 1.012 g of the solute and 15.280 g of the solvent, determine its freezing point in degrees Celsius.

Answer 1

Lets calculate molality first

mass(solute)= 1.012 g

number of mol of solute,

n = mass/molar mass

=(1.012 g)/(78.5 g/mol)

= 0.0129 mol

m(solvent)= 15.280 g

= 0.01528 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(0.0129 mol)/(0.01528 Kg)

= 0.8437 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 6.33*0.8437

= 5.3406 oC

This is decrease in freezing point

freezing point of pure liquid = 6.35 oC

So, new freezing point = 6.35 - 5.3406

= 1.01 oC

Answer: 1.01 oC