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# [Solved] Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 x 10^-3, Kf = 1.86

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Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 x 10-3, Kf = 1.86

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Molar mass of CH3CH2CHClCOOH

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

= 122.5 g mol-1

∴ No. of moles present in 10 g of CH3CH2CHClCOOH

= $$\frac{10 \ g}{122.5 \ gmol^{-1}}$$

= 0.0816 mol

It is given that 10 g of CH3CH2CHClCOOH is added to 250 g of water

∴ Molality of the solution = $$\frac{0.0186}{250}$$x 1000

= 0.3264 mol kg-1

Let αbe the degree of dissociation of CH3CH2CHClCOOH.

CH3CH2CHClCOOH undergoes dissociation according to the following equation:

CH3CH2CHClCOOH ↔ CH3CH2CHClCOO- + H+

Initial conc.    C mol L-1   0            0

At equilibrium  C(1 - α)   Cα         Cα

∴ Ka = $$\frac{C \alpha . C \alpha}{C(1 - \alpha)}$$

= $$\frac{C \alpha^2}{1 - \alpha}$$

Since αis very small with respect to 1, 1 - α

Now, Ka = $$\frac{C \alpha^2}{1}$$

⇒ Ka = Cα2

⇒ α = $$\sqrt{\frac{K_a}{C}}$$

= $$\sqrt{\frac{1.4 \times 10^{-3}}{0.3264}}$$ (∴ Ka = 1.4 x 10-3)

= 0.0655

Again,

CH3CH2CHClCOOH ↔ CH3CH2CHClCOO- + H+

Initial moles    1   0            0

At equilibrium  1 - α   α         Cα

Total moles of equilibrium = 1 - α+ α+ α

= 1 + α

∴ i = $$\frac{1 + \alpha}{1}$$

= 1 + α

= 1+0.0655

= 1.0655

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

ΔTf = i.Kf m

= 1.0655 x 1.86 K kg mol-1 x 0.3264 mol kg-1

= 0.65 K

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