Calculate the depression in the freezing | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

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 x 10-3, Kf = 1.86 K kg mol-1.

Answer:

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°

 


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Comments

  • Nanditha
  • 2019-08-10 13:46:52

Thank u sir :)


  • Dhruvin
  • 2019-06-22 18:12:14

What the formula exactly in this question


  • Balwan Singh
  • 2019-06-17 14:57:55

Thanks


  • Pitambar
  • 2019-05-26 07:13:23

Thnxx


  • Pitambar
  • 2019-05-26 07:12:47

Thnxx


  • Kapil muni mishra
  • 2018-05-04 09:02:01

Thank u sir


Comment(s) on this Question

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 ....