Question

When 9.45 g of ClCH₂COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH₂COOH is x × 10⁻³. The value of x is ______. (Rounded-off to the nearest integer)

[\[\ce{K_{f(H_2O)}}\] = 1.86 K kg mol⁻¹]

पर्याय

• 35

• 40

• 45

• 50

Answer 1

When 9.45 g of ClCH₂COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH₂COOH is x × 10⁻³. The value of x is 35.

Explanation:

ClCH2COOH	⇌	ClCH2COO−	+	H+
Initial Cone. C		O		O
At equilibrium	(1 − α) C	Cα		Cα

∴ Total no. moles = (1 − α) C + α C + α C = C (1 + α)

∴ Van't Hoff factor (i) = `"Observed C.P"/"Normal C.P"`

= `("C"(1 + α))/"C"`

= (1 + α)

ΔT_f = i K_f × m

⇒ `0.5 = (1 + α) xx 1.86 xx (9.45 xx 1000)/(94.5 xx 500)`

⇒ (1 + α) = `2.5/1.86`

⇒ α = `0.64/1.86`

⇒ α = `32/93`

⇒ α = 0.34

C = `(9.45 xx 1000)/(94.5 xx 500)`

⇒ C = 0.2

∴ K^a = `(α^2"C")/(1 - α)`

= `((0.34)^2 xx 0.2)/(1 - 0.34)`

= 35 × 10⁻³