1 g of a carbonate ( M 2 CO 3 ) on treatment with excess HCl produces 0.01 mol of CO 2 . The molar mass of M 2 CO 3 is ______ g mol - 1 . (Nearest integer) [2023]

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Solution

Q.
1 g of a carbonate $(M_{2} {CO}_{3})$ on treatment with excess HCl produces 0.01 mol of ${CO}_{2}$ . The molar mass of $M_{2} {CO}_{3}$ is ______ g ${mol}^{- 1}$ . (Nearest integer) [2023]

Ans.
(100)

$M_{2} {CO}_{3} + HCl \to 2 MCl + H_{2} O$ $+$ ${CO}_{2}$

$1g 0.02 mole 0.01 mole$

$From principle of atomic conservation of carbon atom:$

$Mole of M_{2} {CO}_{3} \times 1 = Mole of {CO}_{2} \times 1$

$\frac{1}{(MM)} \times 1 = (0.01) \times 1$

$Molar mass of M_{2} {CO}_{3} = 100 g/mol$

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