Correct Answer - `3 xx 10^(-4) "molL"^(-1)`

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

`{:(,2BrCl_((g)),hArr,Br_(2(g)),+,Cl_(2(g))),("Initial conc.",3.3xx10^(-3),,0,,0),("At equilibrium",3.3xx10^(-3)-2x,,x,,x):}`

Now, we can write,

`([Br_(2)][Cl_(2)])/([BrCl]^(2)) = K_(c)`

`rArr (x xx x )/((3.3xx10^(-3)-2x)^(2))=32`

`rArr (x)/(3.3xx10^(-3)-2x)=5.66`

`rArr x = 18.678 xx 10^(-3) = 5.66`

`rArr x = 18.678 xx 10^(-3) - 11.32x`

`rArr 12.32 x = 18.678 xx 10^(-3)`

`rArr x = 1.5 xx 10^(-3)`

Therefore, at equilibrium,

`[BrCl] = 3.3 xx 10^(-3) - (2 xx 1.5 xx 10^(-3))`

`= 3.3 xx 10^(-3) - 3.0 xx 10^(-3)`

`= 0.3 xx 10^(-3)`

`= 3.0 xx 10^(-4) "molL"^(-1)`