18 " mL of " 1.0 M `Br_(2)` solution undergoes complete disproportionation in basic medium to `Br^(The hardness of water in terms of )` and `BrO_(3)^(ɵ)`. Then the resulting solution required 45 " mL of " `As^(3+)` solution to reduce `BrO_(3)^(ɵ)` to `Br^(ɵ)`. `As^(3+)` is oxidised to `As^(5+)` which statements are correct?

Options

A. `Ew (Br_(2)) = (M)/(10)`
B. `Ew(Br_(2)) = (5M)/(3)`
C. Molarity of `As^(+3) = 0.4M`
D. Molarity of `As^(3+) = 0.2M`

Answer

B, C

Explanation

`2e^(-) + Br_(2) rarr 2Br^(-) xx 5`
`Br_(2) rarr 2BrO_(3)^(-) +10e^(-)`
`6Br_(2) rarr 5Br^(-) +2BrO_(3)^(-)`
eq of `Br_(2) = (M)/(2) + (10M)/(6) = (5M)/(3)`
mmoles of `Br_(2) = 18 xx 1 = 18`
So mmoles of `BrO_(3)^(-) = (18)/(6) xx 2 = 6`
mEq of `underset((n=2))(As^(+3)) rarr` mEq. of `underset((n = 6))(BrO_(3)^(-))`
`6e^(-) +BrO_(3)^(-) rarr Br^(-)`
mEq of `As^(3+) -=` mEq of `BrO_(3)^(-)`
`45 xx M xx 2` (n-factor) `-= 6 xx 6`
`:. M_(As^(3+)) = (36)/(90) = 0.4M`