5 g of a sample of brass were dissolved in 1 litre dil. `H_(2)SO_(4)`. 20 mL of this solution were mixed with KI and liberated iodine required 20 mL of 0.0327 N hypo solution for titration. Calculate the amount of copper in the alloy.

Explanation

When brass is extracted with concentrated `H_(2)SO_(4)`, it gives copper sulphate.
`2[Cu+2H_(2)SO_(4)to CuSO_(4)+SO_(2)+2H_(2)O]`
`2CuSO_(4)+4KI to 2K_(2)SO_(4)+2CuI_(2)`
`2CuI_(2) to Cu_(2)I_(2)+I_(2)`
`2Na_(2)S_(2)O_(3)+I_(2) to 2NaI+Na_(2)S_(4)O_(6)`
2moles `Cu -= 1 " mole" I_(2) 0-2` mole hypo
20 mL of solution reacts with 20 mL of 0.0327 N hypo
`therefore 1000 mL` of solution will react with 1000 mL of 0.0327 N hypo
No. of moles of hypo used `=("Mass")/("Molecular mass" (158))`
`=(Exx NxxV)/(1000xx158)`
where, E=158, N=0.327 given, V=1000 mL
`therefore` No. of moles of hypo used `=(158xx0.0327xx1000)/(1000xx158)=0.0327`
No. of moles of Cu=No. of moles of hypo =0.0327 mole
Mass of copper in brass `=0.0327xx63.5=2.07645`
% of copper in brass `=(2.07645)/(5)xx100=41.529%`