हिंदी

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.

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प्रश्न

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]

संख्यात्मक
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उत्तर

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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