If one gram of a metal carbonate gave 0.56 gram of its oxide on heating, the equivalent weight of the metal will be

Let metal carbonate be : $MC{O}_3$ and metal oxide is MO

$MC{O}_3⟶MO+C{O}_2$

1gm 0.56g

Number of equivalents of $MC{O}_3$ ​ is equal to the number of equivalents of MO.

So the oxide is CaO (M.wt. = 56) and metal carbonate is $CaC{O}_3$ (M. wt = 100) metal is Ca (M.wt = 40). Eq wt. of metal ion is $C{a}^{2+}$ its eq. wt $\left(\frac{40}{2}\right)=20g$