KEAM 2012 Physics and Chemistry Paper

Let the metal is M, so the formula of its carbonate is MCO3.
Molar mass of MCO3=x+12+3×16
=(x+60)g/mol
(Let atomic mass of M is x .)
MCO3

Δ

→

MO+CO2
1 mol    1 mol = 22.4L
=22400 mL
∵‌448cc‌‌(448mL)‌‌CO2 is produced from carbonate =2g.
∴22400cc‌‌CO2 will be obtained from carbonate
=‌

2×22400

448

=100g
∴‌100=x+60
x=100−60=40g/mol
Eq. wt. of metal =‌

40

2

=20g equiv ‌−1