10 " mL of " a sample of phenol was diluted with `H_2O `and made up to `1.0L`. 20 " mL of " this solution was treted with 40 mL brominating solution (a mixture of `KBrO_3` and KBr) in dil `H_2SO_4`. Excess of KI was added, and the liberated `I_2` required 15 " mL of " 0.1 M `Na_2S_2O_3` for complete reaction. 25 " mL of " the same brominating solution, on similar treatment required, 20 " mL of " 0.1 M `Na_2S_2O_3`. Calculate the weight of phenol per litre of the original sample.

Explanation

Blank titration:
25 " mL of " brominating solution `-=20mL` of `0.1 Na_2S_2O_3`
`(2S_2O_3^(2-)toS_4O_6^(2-)+2e^(-))`
`x=(2)/(2)=1`
(Molarity`=`Normality)
Normality of brominating mixture
`N_1V_1` (brominating mixture) `=N_2V_2(S_2O_3^(-))`
`N_1xx25=0.1xx1xx20`
`N_1=(0.1xx20)/(25)=(2)/(25)`
Titration of brominating mixture `-=` phenol
m" Eq of "brominating mixture `-=` m" Eq of "phenol
`40 mLxx(2)/(25)N=` Total mEq
Excess of brominating mixture `=mEq` of `S_2O_3^(2-)`
`=15xx0.1xx1=15mEq`
Reacted brominating mixture with phenol
`=40xx(2)/(25)-(15xx0.1)=1.7mEq`
m" Eq of "`Br_2=1.7`
m " mol of "`Br_2=(1.7)/(2)(Br_2+2e^(-)+2Br^(ɵ))`
1 mol `-=` 3 " mol of "`Br_2`
m" mol of "phenol `=(1)/(3)xx(1.7)/(2)`
`=(1)/(3)xx(1.7)/(2)xx10^(-3)xx94(g)/(20mL)` ltBrgt `-=(1.7)/(6)xx10^(-3)xx94xx50(g)/(10mL)` of original
`=[(1.7)/(6)xx10^(-3)xx94xx50]100gL^(-1)-=133.16gL^(-1)`
`133.16gL^(-1)` of phenol.