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19 5 g of ch2fcooh is dissolved in 500 g of water...

Question:
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.

Answer:

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 :

ΔT_f = K_f 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)

i = α(1+α) / α = 1 +α = 1.0753

Therefore α = 1.0753- 1

= 0.0753

Now the Value of K_a is given as:

K_a= [CH₂FCOO^-][H⁺] / CH₂FCOOH

= (Cα x Cα) / (C (1-α))

= Cα²/ (1-α)

K_a = 0.5 X (0.0753)² / (1-0.0753)

= 0.5 X 0.00567 / 0.09247

= 0.00307 (approx.)

= 3 X 10^-3

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Question:
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.

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Question: 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.

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Question:
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.