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Question
When 19.5 g of F – CH2 – COOH (Molar mass = 78 g mol−1), is dissolved in 500 g of water, the depression in freezing point is observed to be 1°C. Calculate the degree of dissociation of F – CH2 – COOH.
[Given: Kf for water = 1.86 K kg mol−1]
Solution
Given: Mass = 19.5 g
Molar mass = 78 g mol−1
ΔTf = 1°C
Kf = 1.86 K kg mol−1
ΔTf = Kf × Molality × i
Molality = `"Moles of solute"/"Kg of solvent"`
= `(19.5/78)/(500/100)`
= 0.5 M
i = `(Δ"T"_"f")/("K"_"f" xx "Molality")`
= `1/(1.86 xx 0.5)`
i = 1.07
For F – CH2 – COOH ⇒ i = 1 + α
α = i − 1
= 1.07 − 1
α = 0.07
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