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 cone.	(C mol L-1)	0	0
At equilibrium	C(1 - α)	Cα	Cα

∴ K_α = `("C"α."C"α)/("C"(1- α)) ≃ "C"α^2`

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

To calculate van't Hoff factor:

	\[\ce{CH3CH2CHClCOOH -> CH3CH2CHClCOO^- + H^+}\]
Initial moles	1	0	0
Moles at equilibrium	1 − α	α	α

Total moles = 1 + α

i = 1 + 0.065

∴ i = 1.065

ΔT_f = iK_fm = 1.065 × 1.86 × 0.3264

= 0.649 ≈ 0.65°C