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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
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