100 ml of 2M HCl solution completely neutralises 10 g. of a metal carbonate. Then the equivalent weight of the metal is

Given, $\mathrm{Given}, {\mathrm{M}}_{\mathrm{HCl}}=2\mathrm{M} \Rightarrow {\mathrm{N}}_{\mathrm{HCl}}=2\times \mathrm{basicity}=2\times 1=2\mathrm{N}$

For given neutralization reaction, no. of equivalents of HCl = no. of equivalents of metal carbonate

$$\begin{gathered} {\left(\frac{\mathrm{NV}\left(\mathrm{ml}\right)}{1000}\right)}_{\mathrm{HCl}}={\left(\frac{\mathrm{Wt}.}{\mathrm{E}}\right)}_{\mathrm{metal} \mathrm{carbonate}} ; \mathrm{where} \mathrm{E}=\mathrm{Equivalent} \mathrm{weight} \\ \frac{100\times 2}{1000}=\frac{10}{{\mathrm{E}}_{\mathrm{metal}}+{\mathrm{E}}_{{\mathrm{CO}}_3^{-2}}} \\ 0.2=\frac{10}{\mathrm{x}+30} \left(\because {\mathrm{E}}_{{\mathrm{CO}}_3^{-2}}=\frac{\mathrm{Molar} \mathrm{mass}}{2}=\frac{60}{2}\right) \\ 0.2\mathrm{x}+6=10 \\ \mathrm{x}=20 \end{gathered}$$