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Friday, 9 October

JEE Main 2023 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 13 April 2023, Shift 2, Chemistry Q81

Question 81 of 90 in this shift, Chemistry question 21 of 30, Section B.

1g of a carbonate (M2CO3\mathrm{M_2CO_3}) on treatment with excess HCl produces 0.01 mol of CO2\mathrm{CO_2}. The molar mass of M2CO3\mathrm{M_2CO_3} is ______ g mol−1\mathrm{g\,mol^{-1}}. (Nearest integer)

Official answer

100

NTA final key.