Chemistry - Concentration Terms Question with Solution | TestHub

Let the amount of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ present in the mixture be $\mathrm{xg}.{\mathrm{Na}}_2{\mathrm{SO}}_4$ will not react with ${\mathrm{H}}_2{\mathrm{SO}}_4.$ Then $\frac{x}{53}=\frac{20\times 0.1\times 10}{1000}\therefore x=1.06g$ $\therefore$ Percentage of ${\mathrm{Na}}_2{\mathrm{CO}}_3=$ $\frac{1.06\times 100}{1.25}=84.8\%$