Chemistry - Redox Reactions Question with Solution | TestHub

Equivalent weight of metal carbonate $=20+30=50$ $2.5\mathrm{g}$ of metal carbonate $=\frac{2.5}{50}=0.05\mathrm{eq}$. Number of equivalent of ${\mathrm{H}}_2{\mathrm{SO}}_4$ would have reacted $=0.05$ Number of equivalent of ${\mathrm{H}}_2{\mathrm{SO}}_4$ taken $=\frac{100\times 1}{1000}=0.1$ Number of equivalent of ${\mathrm{H}}_2{\mathrm{SO}}_4$ remains unreacted $=0.1-0.05=0.05\mathrm{eq}$. $\therefore$ Number of equivalent of alkali consumed $=0.05\mathrm{eq}$. milli eq. $=$ Normality $\times$ Volume in $\mathrm{mL}$ $\begin{array}{rl}& \therefore 1.0\times V=0.05\times 1000\\ & V=\frac{0.05\times 1000}{1.0}=50\mathrm{mL}\end{array}$