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)to2Br^(ɵ)]xx5`
`underline(Br_(2)to2BrO_(3)^(ɵ)+10e^(-))`
`underline(6Br_(2)to5Br^(ɵ)+2BrO_(3)^(ɵ))`
Ew of `Br_(2)=(M)/(2)+(M)/(10)=(10M)/(6)=(5M)/(3)`
mmoles of `Br_(2)=18xx1=18`
So mmoles of `BrO_(3)^(ɵ)=(18)/(6)xx2=6`
`m" Eq of "underset((n=2))(As^(+3))tomEq . underset((n=6))(of BrO_(3)^(ɵ))`
`6e^(-)+BrO_(3)^(ɵ)toBr^(ɵ)`
`m" Eq of "As^(3+)-=m" Eq of "BrO_(3)^(ɵ)`
`45xxMxx2` ("n factor")`-=6xx6`
`thereforeM_(As^(3+))=(36)/(90)=0.4`M