0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If `K_(f)` for water is 1.86 K kg `mol^(-1)`, the lowering in freezing point of solutio

Question

0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If `K_(f)` for water is 1.86 K kg `mol^(-1)`, the lowering in freezing point of solutio

0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If `K_(f)` for water is 1.86 K kg `mol^(-1)`, the lowering in freezing point of solution will be :
A. 0.56
B. 1.12 K
C. `-0.56 K`
D. `01.12 K`

Monjur Alahi

7 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - b
`HXhArrH^(+)+X^(-)`
For dessociation of electrolyte,
`alpha=((i-1))/((n-1))or0.2=((i-1))/((2-1))`
i=1+0.2=1.2
`DeltaT_(f)=I K_(f)m`
`=(1.2)xx(1.86 K m^(-1))xx(0.5m)=1.12 K`

7 views Answered 2023-02-05 15:29