1g of a mixture of carbonates of calcium and magnesium gave 240 cm³ of CO₂ at STP .Calculate the percentage composition of the mixture.

mass of CaCO₃=x g and mass of MgCO₃=(1-x) g

CaCO3 gives Cao + CO2

as 100g of CaCO3 fives 22400cm3 of CO2, therefore xg will give 22400/100*x=224x cm3

similarly, 84 g MgCO3 gives 22400 cm3 CO2, Therefore (1-x) g will give 22400/84*(1-x)=800/3*(1-x) cm3

Now, 224x+800/3*(1-x)=240. 

After solving x=5/8.

% of CaCO3= 5/8*100=62.5

% of MgCO3=100-62.5=37.5