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3.0 g sample of KOCl and CaOCl_(2) is dissolved in water to prepare 100 mL solution, which requried 100 " mL of " 0.15 M acidified K_(2)C_(2)O_(4). For the point. The clear solution is now treated with excess of AgNO_(3) solution which precipitates 2.87 g of AgCl. Calculate the mass percentage of KOCl and CaOCl_(2) in the mixture.

Answer» <html><body><p></p>Solution :(a). `2e^(-)+undersetunderset(x=+1)(x-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>=-1)(overset(+1)(ClO^(ɵ))toundersetx=-1)(Cl^(ɵ))(n=2)` <br/> (b). `2e^(-)+undersetnderset(2x=0)(2x-2=-2)(Cl_(2)O^(-2))tounderset(2x=-2)(Cl^(ɵ))(n=2)` <br/> (c). `C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)(n=2)` <br/> Let a and b millimoles of KOCl and `CaCOCl_(2)` are present in the mixture. <br/> `m" Eq of "KOCl+m" Eq of "CaOCl_(2)=m" Eq of "K_(2)C_(2)O_(4)` <br/> `2a+2b=100xx0.15xx2` (n-factor) <br/> `therefore2a+2b=30`.(i) <br/> <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> millomoles of `Cl^(ɵ)` from `KOCl+` millimoles of `Cl^(ɵ)` from `CaOCl_(2)-=` millimoles of AgCl` <br/> `a(lCl^(ɵ)` ions) `+2b(2Cl^(ɵ)` ions)`=(2.87)/(143.5)xx10^(3)` <br/> `[<a href="https://interviewquestions.tuteehub.com/tag/mw-550075" style="font-weight:bold;" target="_blank" title="Click to know more about MW">MW</a> of AgCl=143.5]` <br/> `thereforea+2b=20`(ii) <br/> From <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (i) and (ii) `a=10,b=5` <br/> `% of KOCl=(10xx10^(-3)xx90.5xx100)/(3)` `[Mw of KOCl=90.5]` <br/> `=30.1%` <br/> `% of CaOCl_(2)=(5xx10^(-3)xx127)/(3)xx100``[Mw of CaOCl_(2)=127]` <br/> `=21.1%`</body></html>


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