Question

Calculate the freezing point of a solution that contains 5.42 g of glucose, a non-electrolyte, dissolved in 102 g of water.

Answer 1

Lets calculate molality first

Molar mass of C6H12O6,

MM = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass(C6H12O6)= 5.42 g

use:

number of mol of C6H12O6,

n = mass of C6H12O6/molar mass of C6H12O6

=(5.42 g)/(180.156 g/mol)

= 3.009*10^-2 mol

m(solvent)= 102 g

= 0.102 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(3.009*10^-2 mol)/(0.102 Kg)

= 0.295 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*0.295

= 0.5486 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 0.5486

= -0.5486 oC

Answer: -0.549 oC