Question

Assume 17.65 gram of an ionic substance, MgCl2, was mixed with 125.3 g of acetic acid (CH3COOH). At what temperature will the solution begin to freeze? (the freezing point of acetic acid is 16.6 celsius with a freezing depression constant of 3.59 c/m).

Answer 1

Lets calculate molality first

Molar mass of MgCl2,

MM = 1*MM(Mg) + 2*MM(Cl)

= 1*24.31 + 2*35.45

= 95.21 g/mol

mass(MgCl2)= 17.65 g

number of mol of MgCl2,

n = mass of MgCl2/molar mass of MgCl2

=(17.65 g)/(95.21 g/mol)

= 0.1854 mol

m(solvent)= 125.3 g

= 0.1253 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(0.1854 mol)/(0.1253 Kg)

= 1.479 molal

lets now calculate ΔTf

ΔTf = i*Kf*m

= 3.0*3.59*1.479487

= 15.93 oC

This is decrease in freezing point

freezing point of pure liquid = 16.6 oC

So, new freezing point = 16.6 - 15.93

= 0.67 oC

Answer: 0.67 oC