Question

$5$g of mixture of $NaCl$ and $N{a}_{2}C{O}_3$ was dissolved in water and volume was made $250$ml. Now $25$ml of this solution required $40$ ml of $\frac{N}{10}$ $HCl$ for complete reaction. Find percentage of $NaCl$ in mixture.

Answer 1

Let $a$ be mass of $NaCl$ in the mixture,

mass of $N{a}_{2}C{O}_3=(5-a)g$

In $25ml$, $meqofN{a}_{2}C{O}_3=meqofHCl$

$\therefore meqofN{a}_{2}C{O}_3=40\times \frac{1}{10}=4$

$\therefore$ meq of $N{a}_{2}C{O}_3$ in $250ml=40$ .

$\therefore \frac{\frac{5-a}{106}}{2}\times 1000=40$

$\therefore a=2.88g$

$\therefore \%NaCl$ in the mixture $=\frac{2.88}{5}\times 100=57.6\%$