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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. - CBSE (Science) Class 12 - Chemistry

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Question

Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10−3K= 1.86 K kg mol−1.

Solution

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  = (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 = `0.0186/250 xx 100`

= 0.3264 mol kg-1

Let α be the degree of dissociation of CH3CH2CHClCOOH   

CH3CH2CHClCOOH  undergoes dissociation according to the following equation: CH3CH2CHClCOOH  ↔ CH3CH2CHClCOO- +H+

Initial cone          Cmol L-1           0                       0

At equilibrium     C(1-α)              Cα                    Cα

`:.K_alpha = (Calpha.Calpha)/(C(1-alpha)`

`= (Calpha^2)/(1-alpha)`

Since α is very small with respect to 1, 1 − α ≈ 1

Now `K_alpha = (Calpha^2)/1`

`=>K_alpha = Calpha^2`

`>alpha = sqrt((K_alpha)/C)`

`= sqrt((1.4xx10^(-3))/0.3264)`   (∵ Kα = 1.4 x 10-3)

= 0.0655

Again

                        CH3CH2CHClCOOH  ↔ CH3CH2CHClCOO- +H+

Initial cone            1                                  0                       0

At equilibrium       1-α                             α                        α

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

= 1 + α

`:.i = (1+alpha)/1`

=1+a

=1+0.0655

= 1.0655

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

`triangleT_f = i.K_fm`

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

= 0.65 K

  Is there an error in this question or solution?

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Solution 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. Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point.
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