Question

19.5 gram of ch 2 ch is dissolved in 500 gram of water is the depression in freezing point of water is absorbed 81

Answer 1

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W₁ = 500 g 

W₂ = 19.5 g 

K(f) = 1.86 K kg mol⁻¹

ΔT(f) = 1K

As we know ,

72.54 g mol⁻¹

I.e.       Observed mass of CH₂FCOOH is  72.4 g mol⁻¹

Calculated molar mass of CH₂FCOOH is = 14 + 19 + 12 + 16 + 16 + 1 

                                                                  = 78 g mol⁻¹

Van't Hoff Factor 

78 / 72 . 54 = 1.0753

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Let α Be the degree of disassociation 

CH₂FCOOH            ⇄     CH₂FCOO⁻    +       H⁺

Initial concentration             0                          0

At Equilibrium                      Cα                       Cα

Total = C(1+α)

∴ 

i = 1 + α 

α = i - 1 

   = 1.0753 - 1 

    

   = 0.0753

Value of K(α) is given is 

By Volume of the solution  as 500 mL 

The concentration will be :

   

          = 0.5 M 

Therefore, 

K(α)       = 0.00307 (aprox.)

  

              = 3. 07 X 10⁻³

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