Calculate the depression in the freezing point of water when $10 g$ of $C H_{3} C H_{2} C H C l C O O H$ is added to $250 g$ of water.
[ $K a = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}$ ]

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Solution

The molar mass of $2$ -chloro butanoic acid is $122.5$ g/mol.

Number of moles $= \frac{10}{122.5} = 0.0816$ mol

Molality of solution $= \frac{0.0816 \times 1000}{250} = 0.3265$ m

Let $α$ be the degree of dissociation and c be the initial concentration. The concentration after dissociation is as shown.
$C H_{3} C H_{2} C H C l C O O H c (1 - α) ⇌ C H_{3} C H_{2} C H C l C O O^{-} c α + {H c α}^{+}$

The equilibrium constant expression is

$K = \frac{c α \times c α}{c (1 - α)} = c α^{2}$

$α = \sqrt{\frac{K}{c}} = \sqrt{\frac{1.4 \times 10^{- 3}}{0.3265}} = 0.065$

Calculation of vant Hoff factor:

$C H_{3} C H_{2} C H C l C O O H (1 - α) ⇌ C H_{3} C H_{2} C H C l C O O^{-} α + α H^{+}$

$i = \frac{1 - α + α + α}{1} = 1 + α = 1 + 0.065 = 1.065$

The depression in the freezing point $Δ T_{f} = i K_{f} m = 1.065 \times 1.86 \times 0.3265 = {0.647}^{o}$

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Question 2.32

Calculate the depression in the freezing point of water when 10g of $C H_{3} C H_{2} C H C l C O O H$ is added to 250 g of water, $K_{a} = 1.4 \times 10^{- 3}$ ; $K_{f} = 1.86 K k g m o l^{- 1}$

Q. Calculate the depression in freezing point of water when

20.0 g

C H_{3} C H_{2} C H C l C O O H

is added to

500 g

of water.
[Given:

K_{a} = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}]

Q. Calculate the amount of

K C l

which must be added to 1 kg of water so that the freezing point is depressed by 2K.

(k_{f}

for water

= 1.86 K k g m o l^{- 1})

Q. Calculate the freezing point of a solution containing 60g of glucose (Molar mass = 180 g

M o l^{- 1})

in 250 g of water.

(k_{f} o f w a t e r = 1.86 K k g m o l^{- 1})

Q. (i) The depression in freezing point of water observed for the same molar concentration of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order as stated above. Explain.
(ii) Calculate the depression in freezing point of water when 20.0 g of

C H_{3} C H_{2} C H C l C O O H

is added to 500g of water.

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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=1.4×10−3,Kf=1.86 K kg mol−1]

Calculate the depression in the freezing point of water when $10 g$ of $C H_{3} C H_{2} C H C l C O O H$ is added to $250 g$ of water.
[ $K a = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}$ ]