Question

$3.0$ g sample of $KOCl$ and $CaOC{l}_2$ is dissolved in water to prepare $100$ mL solution which requires $100$ mL of $0.15M$ acidified $K_2{C}_2{O}_4$ for end point. The clear solution is now treated with excess of $AgN{O}_3$ solution which precipitates $2.87$ g of $AgCl$. The mass percentage of $KOCl$ and $CaOC{l}_2$ in the mixture is :

• A. $21.9$%
• B. $21.5$%
• C. $21.1$%
• D. none of the above

Answer 1

Answer: (C)

$21.1$%

The redox changes involved are shown below:
$a$. $2e^{-}+\stackrel{+1}{\underset{\underset{x=+1}{x-2=-1}}{C1O^{⊝}}}\to \underset{x=-1}{C{1}^{⊝}}(n=2)$
$b$. $2e^{-}+\underset{\underset{2x=0}{2x-2=-2}}{C{1}_2{O}^{-2}}\to \underset{2x=-2}{C{1}^{⊝}}(n=2)$
$c$. $C_2{O_4}^{2-}\to 2C{O}_2+2e^{-}(n=2)$
Let $a$ and $b$ millimoles of $KOCl$ and $CaOC{l}_2$ are present in the mixture respectively.
$mEq$ of $KOCl+mEq$ of $CaOC{l}_2=mEq$ of $K_2{C}_2{O}_4$
$(n=2)$ $(n=2)$ $(n=2)$
$2a+2b=100\times 0.15\times 2(n$-factor$)$
$\therefore 2a+2b=30$ $.......\left(i\right)$
Also, millimoles of $C{l}^{⊝}$ from $KOCl$ + millimoles of $C{l}^{⊝}$.from $CaOC{l}_2\equiv$ millimoles of $AgCl$
$a\left(1C{l}^{⊝}ions\right)+2b\left(2C{l}^{⊝}ions\right)=\frac{2.87}{143.5}\times {10}^3$ $[Mw$ of $AgC1=143.5]$
$\therefore a+2b=20$ $.......\left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$, $a=10,b=5$.
Total weight of sample $=3$ g.
% of $KOCl=\frac{10\times {10}^{-3}\times 90.5\times 100}{3}$
$[Mw$ of $KOCl=90.5]$ $=30.1$%
% of $CaOC{l}_2=\frac{5\times {10}^{-3}\times 127}{3}\times 100$
$[Mw$ of $CaOC{l}_2=127]$ $=21.1$%