1g of a carbonate (M2CO3) on treatment with excess HCl produces 0.01 mol of CO2. The molar mass of M — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

M2CO3 + HCl  2MCl + H2O + CO2 1g 0.02 mole 0.01 mole POAC on C-atom Mole of M2CO3 × 1 = Mole of CO2 × 1    1 . 1 M . M    molar mass of M2CO3 = 100 g/mol