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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.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> of sodium thiosulphate solution may be determined as : <br/> `N_(1)V_(1)(Na_(2)S_(2)O_(3))=N_(2)V_(2) ("pure" K_(2)Cr_(2)O_(7))` <br/> `N_(1)xx26=25xx(1)/(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)` <br/> `N_(1)=0.048` (<a href="https://interviewquestions.tuteehub.com/tag/hypo-1034629" style="font-weight:bold;" target="_blank" title="Click to know more about HYPO">HYPO</a>) <br/> The <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> involved may be given as : <br/> `Cr_(2)O_(7)^(2-)+6O^(-)+4H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O` <br/> `3[I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(4)O_(6)]` <br/> 1 mole `K_(2)Cr_(2)O_(7)-=6 " mole" Na_(2)S_(2)O_(3)` <br/> 25 mL of solution of `K_(2)Cr_(2)O_(7)` is treated by 24 mL of 0.048 N hypo <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> 500 mL` of solution will be titrated by 480 mL of 0.048 N hypo <br/> No. of moles of hypo`=("Mass")/(M.w. (158))=(ExxNxxV)/(1000xx158)` <br/> `=(158xx0.048xx480)/(1000xx158)` <br/> =0.02304 mole <br/> No. of moles of `K_(2)Cr_(2)O_(7)=(1)/(6)`[No. of moles of hypo] <br/> `=(1)/(6)[0.02304]=3.84xx10^(-3)` <br/> Mass of `K_(2)Cr_(2)O_(7)=3.84xx10^(-3)xx294=1.12896` <br/> % purity `=(1.12896)/(1.5)xx100=75.26%`</body></html>


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