Question

Calculate the depression in the freezing point of water when 10g

Answer 1

Answer:

Molar mass of CH3CH2CHClCOOH

15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1

= 122.5 g/mol

∴ Moles of CH3CH2CHClCOOH = 10g / 122.5 g/mol

= 0.0816 mol

Therefore molality of the solution

= (0.0816 x 1000) / 250

= 0.3265 mol kg-1

Now if a is the degree of dissociation of CH3CH2CHClCOOH,

So, Ka = (Cα x Cα) / (C (1-α))

Ka = Cα2 / (1-α)

Since α is very small with respect to 1, 1 - α = 1

Ka = Cα2

α = √Kα / C

Putting the values ,We get

α = √1.4 x 10-3 / 0.3265

= 0.0655

Now at equilibrium,the van’t hoff factor i = 1-α +α +α/1

= 1 + 0.0655

= 1.0655

Hence, the depression in the freezing point of water is given as:

Therefore ΔTf = i Kf m v

= 1.065 v x 1.86 x 0.3265

= 0.647°

Explanation: