1.5 g of sample of impure potassium dichromate was dissolved in water and made up to 500 mL solution . 25 mL of this solution required iodometrically 24 mL of a sodium thiosulphate solution. 26 mL of this sodium thisulphate solution required 25 mL of `N//20` solution of pure potassium dichromate. Find the percentage purity of impure sample of potassium dichromate.

Explanation

Normality of sodium thiosulphate solution may be determined as :
`N_(1)V_(1)(Na_(2)S_(2)O_(3))=N_(2)V_(2) ("pure" K_(2)Cr_(2)O_(7))`
`N_(1)xx26=25xx(1)/(20)`
`N_(1)=0.048` (hypo)
The reaction involved may be given as :
`Cr_(2)O_(7)^(2-)+6O^(-)+4H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O`
`3[I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(4)O_(6)]`
1 mole `K_(2)Cr_(2)O_(7)-=6 " mole" Na_(2)S_(2)O_(3)`
25 mL of solution of `K_(2)Cr_(2)O_(7)` is treated by 24 mL of 0.048 N hypo
`therefore 500 mL` of solution will be titrated by 480 mL of 0.048 N hypo
No. of moles of hypo`=("Mass")/(M.w. (158))=(ExxNxxV)/(1000xx158)`
`=(158xx0.048xx480)/(1000xx158)`
=0.02304 mole
No. of moles of `K_(2)Cr_(2)O_(7)=(1)/(6)`[No. of moles of hypo]
`=(1)/(6)[0.02304]=3.84xx10^(-3)`
Mass of `K_(2)Cr_(2)O_(7)=3.84xx10^(-3)xx294=1.12896`
% purity `=(1.12896)/(1.5)xx100=75.26%`