50.0 g sample of brass is dissolved in 1 L dil `H_2SO_4.20mL` of this solution is mixed with KI, and the liberated `I_2` required 20 " mL of " 0.5 M hypo solution for titration calculate the amount of Cu in the alloy.

Explanation

`Cu-=CuSO_4-=KI-=I_2-=Na_2S_2O_3`
`mEq-=mEq-=mEq-=mEq`
`?-=--=--=20xx0.5xx1(n=1)`
`=(10 m" Eq of "Na_2S_2O_3)/(20 mL "of solution")`
`-=(10xx1000)/(20)mEqL^(-1)=500mEqL^(-1)` of solution
m" Eq of "Cu `=500xx10^(-3)xx(63.5)/(1)` (n-factor `=1`)
`=(635xx5)/(100)=31.75g`
`%` `Cu=(31.75)/(50)xx100=63.5%`