Question

icalculate the freezing temperature of solution if KBr is 80% dissociated in aqueous solution of 0.5m concentration.

Answer 1

hope it may help you

Answer 2

Answer:

271.326K

Explanation:

here

∆ = 0.r

n=4. {KBr ---- ionic

i.e K^+ ,. Br^-}

= ∆ (n-1) +1

= 0.8 +1

= 1.8

∆Tf = L. K. f x m

= 1.8. 1.86. 0.5

= 1.65

freezing point = 273-1.65

= 271.326

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