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25 mL of an aqueous solution of KCl was ...

`25 mL` of an aqueous solution of `KCl` was found to requires `20 mL` of `1M AgNO_(3)` solution when titrated using a `K_(2)CrO_(4)` as indicator. Depression in freezing point of `KCl` solution with `100%` ionisation will be :
`(K_(f) = 2.0 mol^(-1) kg "and molarity = molality")`

A

(a) `5.0`

B

(b) `3.2`

C

(c ) `1.6`

D

(d) `0.8`

Text Solution

Verified by Experts

The correct Answer is:
B
milli-mole of `KCl = m-"mole of" AgNO_(3)`
`25 xx M = 20 xx 1`
`:. M_(KCl) = (20)/(25)`
Now `Delta T = K_(f) xx " molality"(1+alpha)`
`= 2 xx (20)/(25) (1+1) = (80)/(25) = 3.2`
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