Question

Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. K_a = 1.4 × 10⁻³, K_f = 1.86 K kg mol⁻¹.

Answer 1

Molar mass of CH₃CH₂CHClCOOH = 122.5 g mol⁻¹

∴ No. of moles present in 10 g of CH₃CH₂CHClCOOH = `10/122.5` mol

= 8.16 × 10⁻² mol

Molality of the solution (m) = `(8.16 xx 10^-2)/250 xx 1000` = 0.3264

Let α be the degree of dissociation of CH₃CH₂CHClCOOH   

CH₃CH₂CHClCOOH undergoes dissociation according to the following equation:

	\[\ce{CH3CH2CHClCOOH -> CH3CH2CHClCOO- + H+}\]
Initial conc.	(C mol L−1)                                     0                     0
At equilibrium	C(1 − α)                                         Cα                 Cα

∴ K_α = `(C alpha * C alpha)/(C(1 - α)) ≃ Cα^2`

or, α = `sqrt (K_α//C) = sqrt ((1.4 xx 10^-3)/0.3264)` = 0.065

To calculate the van’t Hoff factor:

	\[\ce{CH3CH2CHClCOOH -> CH3CH2CHClCOO- + H+}\]
Initial conc.	1                                          0                         0
At equilibrium	(1 − α)                                     α                         α

Total moles = 1 + α

i = 1 + 0.065

∴ i = 1.065

ΔT_f = i K_f m

= 1.065 × 1.86 × 0.3264

= 0.649 ≈ 0.65°C