1 gram of a metal carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of (M2CO3) in gmol−1 is
A
1186
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B
84.3
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C
118.6
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D
11.86
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Solution
The correct option is B84.3 M2CO3+2HCl→2MCl+H2O+CO2 0.01186 moles CO2=0.01186 moles ofM2CO3=1gM2CO3 Molar mass of M2CO3=Mass of M2CO3No. of moles of M2CO3=1g0.01186mol=84.3g/mol