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20 mL of a solution containing ferrous sulphate and ferric sulphate acidified with H_(2)SO_(4) is reduced by metallic zinc. The solution required 27.4 mL of 0.1 N solution of K_(2)Cr_(2)O_(7) for oxidation. However before reduction with zinc, 20 mL of same solution required 17.96 mL of same K_(2)Cr_(2)O_(7). Calculate the mass of FeSO_(4) " and " Fe_(2)(SO_(4))_(3) per litre of the solution.

Answer» <html><body><p><br/></p>Solution :After reduction `Fe_(2)(SO_(4))_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` is also reduced to `FeSO_(4)` and titration <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> total concentration of `FeSO_(4) " and" Fe_(2)(SO_(4))_(3)`. <br/> Titration before reduction gives only `FeSO_(4)`. <br/> Milli <a href="https://interviewquestions.tuteehub.com/tag/equiv-2618698" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIV">EQUIV</a>. of `K_(2)Cr_(2)O_(7)` after reduction `=27.4xx0.1=2.740` <br/> Milli equiv. of `K_(2)Cr_(2)O_(7)` before reduction `=17.96xx0.1=1.796` <br/> Milli equiv. of `Fe_(2)(SO_(4))_(3)` in 20 mL =0.944 <br/> Milli equiv. of `FeSO_(4)` in 20 mL =1.796 <br/> `therefore FeSO_(4)(gL^(-1))=(1.796)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)xx"Eq. mass"xx40` <br/> `=(1.796)/(1000)xx152xx40=10.92` <br/> `Fe_(2)(SO_(4))_(3)(gL^(-1))=(0.944)/(1000)xx"Eq. mass"xx40` <br/> `=(0.944)/(1000)xx200xx40=7.55`</body></html>


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