4.2 g of a metallic carbonate MCO₃ was heated in a hard glass tube and CO₂ evolved was found to have 1120 mL of volume at STP. The Ew of the metal is

At STP conditions, 22400 mL equivalent to 1 mol of a gas. Therefore, it can be said that one mole of CO₂ has 22400 mL of volume.

Number of moles of CO₂ present in 1120 mL of CO₂ is calculated as:

$\begin{array}{ccc}22400 \mathrm{mL}& =& 1 \mathrm{mol}\\ 1120 \mathrm{mL}& =& 1120 \mathrm{mL}\times \frac{1 \mathrm{mol}}{22400 \mathrm{mL}}\\ & =& 0.05 \mathrm{mol}\end{array}$

On Heating, one mole of metal carbonate (MCO₃) forms one mole of CO₂ and one mole of metal oxide (MO).

${\mathrm{MCO}}_3\to {\mathrm{CO}}_2+\mathrm{MO}$

From the above reaction, it also concludes that 0.05 mol of CO₂ forms from 0.05 mol of MCO₃. 

Assume that atomic weight of M (metal) is x.

$\begin{array}{ccc}\mathrm{Molecular} \mathrm{weight} \mathrm{of} {\mathrm{MCO}}_3& =& \mathrm{Atomic} \mathrm{weight} \mathrm{of} \mathrm{M}+\mathrm{Atomic} \mathrm{weight} \mathrm{of} \mathrm{C}+3(\mathrm{atomic} \mathrm{weight} \mathrm{of} \mathrm{O})\\ & =& \mathrm{x}+12+3\left(16\right)\\ & =& \mathrm{x}+60\end{array}$

Formula that shows relation between the number of moles, the given mass, and the molecular weight is:

$\mathrm{Number} \mathrm{of} \mathrm{moles}=\frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Molecular} \mathrm{weight}}$

Value of x in x+60 is calculated by substituting the given values in the above expression.

$\begin{array}{ccc}0.05 \mathrm{mol}& =& \frac{4.2 \mathrm{g}}{\mathrm{x}+60}\\ \mathrm{x}& =& 24\end{array}$

Atomic weight of metal is 24. Equivalent weight of metal is determined by diving the atomic weight of element by its valency. 

$\mathrm{Equivalent} \mathrm{weight}=\frac{\mathrm{Atomic} \mathrm{weight}}{\mathrm{Valency}}$

The given formula of metal carbonate determines that the valency of metal is 2.

${\mathrm{MCO}}_3\to {\mathrm{M}}^{2+}+{\mathrm{CO}}_3^{2-}$

Hence, equivalent weight of metal (M) is calculated as:

$\begin{array}{ccc}\mathrm{Equivalent} \mathrm{weight}& =& \frac{24}{2}\\ & =& 12\end{array}$