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 K kg mol-1.
Molar mass of CH3CH2CHClCOOH
15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1
= 122.5 g/mol
∴ Moles of CH3CH2CHClCOOH = 10g / 122.5 g/mol
= 0.0816 mol
Therefore molality of the solution
= (0.0816 x 1000) / 250
= 0.3265 mol kg-1
Now if a is the degree of dissociation of CH3CH2CHClCOOH,
So, Ka = (Cα x Cα) / (C (1-α))
Ka = Cα2 / (1-α)
Since α is very small with respect to 1, 1 - α = 1
Ka = Cα2
α = √Kα / C
Putting the values ,We get
α = √1.4 x 10-3 / 0.3265
= 0.0655
Now at equilibrium,the van’t hoff factor i = 1-α +α +α/1
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
Therefore ΔTf = i Kf m v
= 1.065 v x 1.86 x 0.3265
= 0.647°
The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate of formation of D/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
6.0 × 10 - 3 |
| II | 0.3 | 0.2 |
7.2 × 10 - 2 |
| III | 0.3 | 0.4 |
2.88 × 10 - 1 |
| IV | 0.4 | 0.1 |
2.40 × 10 - 2 |
Determine the rate law and the rate constant for the reaction.
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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 32: Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 ....
Comments
Thank u sir :)
What the formula exactly in this question
Thanks
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Thnxx
Thank u sir