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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

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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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.

बेरीज
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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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2025-2026 (March) Model set 1 by shaalaa.com

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