[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
Calculate the depression in the freezing point of water when 10 g of CH_{3}CH_{2}CHClCOOH is added to 250 g of water. K_{a} = 1.4 x 10^{3}, K_{f} = 1.86
Molar mass of CH_{3}CH_{2}CHClCOOH
= 15+14+13+35.5+12+16+16+1
= 122.5 g mol^{1}
∴ No. of moles present in 10 g of CH_{3}CH_{2}CHClCOOH
= \(\frac{10 \ g}{122.5 \ gmol^{1}}\)
= 0.0816 mol
It is given that 10 g of CH_{3}CH_{2}CHClCOOH 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 CH_{3}CH_{2}CHClCOOH.
CH_{3}CH_{2}CHClCOOH undergoes dissociation according to the following equation:
CH_{3}CH_{2}CHClCOOH ↔ CH_{3}CH_{2}CHClCOO^{} + H^{+}
Initial conc. C mol L^{1} 0 0
At equilibrium C(1  α) Cα Cα
∴ K_{a} = \(\frac{C \alpha . C \alpha}{C(1  \alpha)}\)
= \(\frac{C \alpha^2}{1  \alpha}\)
Since αis very small with respect to 1, 1  α
Now, K_{a} = \(\frac{C \alpha^2}{1}\)
⇒ K_{a} = Cα^{2}
⇒ α = \(\sqrt{\frac{K_a}{C}}\)
= \(\sqrt{\frac{1.4 \times 10^{3}}{0.3264}}\) (∴ K_{a} = 1.4 x 10^{3})
= 0.0655
Again,
CH_{3}CH_{2}CHClCOOH ↔ CH_{3}CH_{2}CHClCOO^{} + 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:
ΔT_{f} = i.K_{f} m
= 1.0655 x 1.86 K kg mol^{1} x 0.3264 mol kg^{1}
= 0.65 K

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