Question

$0.5$ g of a mixture of $K_{2}C{O}_3$ and $L{i}_{2}C{O}_3$ requires $30$ mL of $0.25$ N $HCl$ solution for neutralization. The percentage composition of the mixture is:

• A. $96\%K_{2}C{O}_3,4\%L{i}_{2}C{O}_3$
• B. $4\%$ $K_{2}C{O}_3$, $96\%$ $L{i}_{2}C{O}_3$
• C. $48\%$ $K_{2}C{O}_3$, $52\%$ $L{i}_{2}C{O}_3$
• D. $52\%$ $K_{2}C{O}_3$, $48\%$ $L{i}_{2}C{O}_3$

Answer 1

The correct option is A $96\%K_{2}C{O}_3,4\%L{i}_{2}C{O}_3$

Molecular weight of $K_{2}C{O}_3=39\times 2+12+48=138g$ $mo{l}^{-1}$
Equivalent weight of $K_{2}C{O}_3=\frac{138}{2}=69g$
Molecular weight of $L{i}_{2}C{O}_3=7\times 2+12+48=74g$ $mo{l}^{-1}$
Equivalent weight of $L{i}_{2}C{O}_3=\frac{74}{2}=37g$
Let, $x$ g of $K_{2}C{O}_3$ and $(0.5-x)$ g of $L{i}_{2}C{O}_3$
mEq $K_{2}C{O}_3$+ mEq of $L{i}_{2}C{O}_3$= mEq of $HCl$
$\left(\frac{x}{69}+\frac{0.5-x}{37}\right)\times 1000=30\times 0.25$
$\frac{37x+69(0.5-x)}{69\times 37}=\frac{30\times 0.25}{1000}$
$69\times 0.5=\frac{30\times 0.25\times 69\times 37}{1000}=32x$
$34.5-19.14=32x$
$32x=15.36$ and $x=0.48$
$\%$ of $K_{2}C{O}_3=\frac{0.48\times 100}{0.5}=96\%$
$\%$ of $L{i}_{2}C{O}_3=4\%$