Volume of CO₂ evolved = 1336 mL, Temperature = 273 + 27 = 300 K, Pressure = 700 mm

Converting the volume of CO₂ to volume at STP by using the gas equation,

$${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$$ ; $${{700 \times 1336} \over {300}} = {{760 \times V} \over {273}}$$

V = 1120 mL

Where V is the volume of CO₂ at STP.

Now, 22400 mL of CO₂ at STP weigh = Gram mol. wt. of CO₂ = 44 g

$$\therefore$$ 1120 mL of CO₂ at STP weigh

$$ = {{44} \over {22400}} \times 1120 = 2.20$$ g

Weight of metal carbonate = 4.215 g

Weight of CO₂ evolved = 2.20 g

Weight of metal oxide left after evolution of CO₂ gas = 4.215 $$-$$ 2.20 = 2.015 g

If equivalent weight of metal is E then using the relation:

$${{Weight\,of\,metal\,carbonate} \over {weight\,of\,metal\,oxide}}$$

$$ = {{E + Eq.\,wt.\,of\,C{O_3}^{2 - }} \over {E + Eq.\,wt.\,of\,{O^{2 - }}}}$$

Substituting values $${{4.215} \over {2.015}} = {{E + 30} \over {E + 8}}$$ [$$\because$$ Eq. wt. of $$C{O_3}^{2 - } = {{60} \over 2} = 30$$]

On solving, E = 12.15