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25.0 cm^(3) of a solution containing 15.0 g of a partially oxidisied sample of green vitrion (FeSO_(4).7H_(2)O) per litre required 20.0 cm^(3) mL of 0.01 M potassium dichromate solution for oxidiation in aidic medium find out the percentage purity of the given sample of green vitriol

Answer» <html><body><p></p>Solution :Step 1 To write balaced <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> for the redox reaction <br/> `K_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)Cr_(2)+<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> H_(2)SO_(4) rarr K_(2)SO_(4)+Cr_(2)(sO_(4))_(3)+4H_(2)O+3O` <br/> `2FeSO_(4)+H_(2)SO_(4)+OrarrFe_(2)(SO_(4))+H_(2)O[xx3` <br/> `K_(2)Cr_(2)O_(7)+6FeSO_(4)+7H_(2)SO_(4)rarr K_(2)SO_(4)+Cr_(2)SO_(4)^(3)+3Fe_(2)SO_(4)^(3)+7H_(2)O` <br/> From the above equation <br/> 1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of `K_(2)Cr_(2)O_(7)=6` <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of `FeSO_(4)` <br/> Step 2 To find percentage purity of green vitroit Let `M_(1)` be the molarity of the oxidised sample of green vitril applying molarity eqaution we have <br/> `(M_(1)xx25)/(6)(FeSO_(4))=(20xx0.1)/(1)(K_(2)Cr_(2)O_(7))` <br/> `M_(1)=(20xx0.01xx6)/(25)=0.048 M` <br/> Mol wt of `FeSO_(4)7H_(2)O=56+32+4xx16+7xx18=278` <br/> Wt of pure `FeSO_(4), 7H_(2)O=278xx0.048=13.344 g L^(-1)` <br/> % purity of green vitriol `=(13.444)/(15)xx100=88.96`</body></html>


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