Question

Determine the freezing point (C⁰) of a solution that contains 25.0g of NaCl in 200g of h2o

Answer 1

Lets calculate molality first

Molar mass of NaCl = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass of NaCl = 25.0 g

we have below equation to be used:

number of mol of NaCl,

n = mass of NaCl/molar mass of NaCl

=(25.0 g)/(58.44 g/mol)

= 0.4278 mol

mass of solvent = 200 g

= 0.2 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(0.4278 mol)/(0.2 Kg)

= 2.139 molal

i for NaCl = 2 as it dissociates into 2 particles

Kf for water = 1.86 oC/molal

lets now calculate deltaTf

deltaTf = i*Kf*m

= 2.0*1.86*2.1389

= 7.9569 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 7.96

= -7.96 oC

Answer: -7.96 oC