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2.0 g sample of NaCN is dissolved in 50 " mL of " 0.3 M mild alkaline KMnO_(4) and heated strongly to convert all the CN^(ɵ) to OCN^(ɵ). The solution after acidification with H_(2)SO_(4) requries 500 " mL of " 0.05 M FeSO_(4) Calculatethe percentage purity of NaCN in the sample.

Answer» <html><body><p></p>Solution :(a). `undersetunderset(x=2)(x-3=-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)(Coverset(+2)(N^(ɵ)))toundersetunderset(x=<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)(x-3-2=-1)(OCoverset(+4)(N^(ɵ)))+<a href="https://interviewquestions.tuteehub.com/tag/2e-300683" style="font-weight:bold;" target="_blank" title="Click to know more about 2E">2E</a>^(-)(n=2)` <br/> (b). `3e^(-)+undersetunderset(x=7)(x-8=-1)(MnO_(4)^(ɵ))toundersetunderset(x=3)(x-4=0)(MnO_(2))`<br/> (c). `5e^+MnO_(4)^(ɵ)toMn^(2+)` `(n=5` in acidic medium) <br/> (d). `Fe^(2+)toFe^(3+)+2e^(-)(n=1)` <br/> `m" Eq of "KMnO_(4)` added in basic medium`=50xx0.3` <br/> (n-factor`=3`) <br/> `=45.0` <br/> `m" Eq of "KMnO_(4)` in acidic medium `(n=5)` <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> after reaction with `NaCN=m" Eq of "FeSO_(4)` used <br/> `=500xx0.5xx1` (n-factor`=1`) <br/> `=25.0` <br/> `m" Eq of "KMnO_(4)` (n-factor`=3`) left`=(25xx3)/(5)=15` <br/> `m" Eq of "NaCN` in sample `=m" Eq of "KMnO_(4)` added `-m" Eq of "KMnO_(4)` left <br/> `=45-15=30` <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>(Weight)/(Ew of NaCN)xx10^(3)=30` <br/> `[Ew of NaCN=(49)/(2)` (n factor`=2`)] <br/> `(W)/((49)/(2))xx10^(3)=30` <br/> `W_(NaCN)=0.735g`. <br/> `% of NaCN=(0.735)/(2.0)xx100=36.75%`</body></html>


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