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
Answers
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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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