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Academic session 2026-27, Chapter 1 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter Solutions. This page offers a step-by-step solution to the specific question from Exercise 2, Question 33:
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
Molecular mass of CH2FCOOH
14 + 19 + 12 + 16 + 16 + 1 = 78 g/mol
Now, Moles of CH2FCOOH = 19.5 / 78
= 0.25
Taking the volume of the solution as 500 mL, we have the concentration:
C = (0.25 / 500) X 1000
Therefore Molality = 0.50m
So now putting the value in the formula :
ΔTf = Kf x m
=1.86 x 0.50 = 0.93K
Van’t hoff factor = observed freezing point depression / calculated freezing point depression
= 1 / 0.93 = 1.0753
Let α be the degree of dissociation of CH2FCOOH

Now total number of moles = m(1-a) + ma +ma = m(1+a)
Or
i = α(1+α) / α = 1 +α = 1.0753
Therefore α = 1.0753- 1
= 0.0753
Now the Value of Ka is given as:
Ka = [CH2FCOO-][H+] / CH2FCOOH
= (Cα x Cα) / (C (1-α))
= Cα2 / (1-α)
Ka = 0.5 X (0.0753)2 / (1-0.0753)
= 0.5 X 0.00567 / 0.09247
= 0.00307 (approx.)
= 3 X 10-3
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 33: 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water obs....
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Comments
Why do we take, K(f) to be 1.86?
Value of c will be equals to molality
Great job
Great job
What a explanation
How the value of C is calculated for Ka...???
C can be calculated by c=n/v
How the value of 'C' is calculated for Ka value.
Great hard work.