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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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Solution
It is given that:
w1 = 500 g
w2 = 19.5 g
Kf = 1.86 K kg mol−1
ΔTf = 1.0°C
We know that:
M2 = `(K_f xx w_2 xx 1000)/(Delta T_f xx w_1)`
= `(1.86 K kg "mol"^(-1) xx 19.5 g xx 1000 g kg^(-1))/(500 g xx 1.0)`
= 72.54 g mol−1
∴ Observed molar mass of CH2FCOOH, (M2)obs = 72.54 g mol−1
The calculated molar mass of CH2FCOOH is (M2)cal = 14 + 19 + 12 + 16 + 16 + 1
= 78 g mol−1
∴ Van’t Hoff factor (i) = `((M_2)_(cal))/(M_2)_(obs)`
= `78/72.54`
= 1.0753
Let α be the degree of dissociation of CH2FCOOH.
| \[\ce{CH2FCOOH ⇌ CH2FCOO^- + H^+}\] | |||
| Initial Conc. | C mol L−1 0 0 | ||
| At equilibrium | C(1− α) Cα Cα | ||
∴ i = `(C(1 + α))/C`
⇒ i = 1 + α
⇒ α = i − 1
= 1.0753 − 1
= 0.0753
Now, the value of Kα is given as:
Kα = `([CH_2FCOO^-][H^+])/([CH_2FCOOH])`
= `(C alpha * C α)/(C (1 - alpha))`
= `(C alpha^2)/(1 - alpha)`
Taking the volume of the solution as 500 mL, we have the concentration:
C = `19.58/78 xx 1/500 xx 1000`
= 0.5 M
∴ Kα = `(C alpha^2)/(1 - alpha)`
= `(0.5 xx (0.0753)^2)/(1 - 0.0753)`
= `(0.5 xx 0.00567)/0.9247`
= 0.00307 (approximately)
= 3.07 × 10−3
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