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प्रश्न
0.01 m aqueous formic acid solution freezes at – 0.021°C. Calculate its degree of dissociation, Kf = 1.86 K kg mol–1.
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उत्तर
ΔTf = iKfm
ΔTf = 0°C – (–0.021°C) = 0.021°C
m = 0.01 mol kg–1
0.021 = i × 1.86 K kg mol–1 × 0.01 mol kg–1
i = `0.021/(1.86 xx 0.01)` = 1.13
α = `(i - l)/(n - l)` = i – l because n = 2
Hence, α = 1.13 – 1 = 0.13 = 13%
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