The reaction between M\(_2\)CO\(_3\) and HCl is: \[ \text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{H}_2\text{O} + \text{CO}_2. \] From the principle of atomic conservation, 1 mole of M\(_2\)CO\(_3\) produces 1 mole of CO\(_2\). Given: \[ \text{Moles of CO}_2 = 0.01 \, \text{mol}. \] \[ \text{Moles of M}_2\text{CO}_3 = 0.01 \, \text{mol}. \] The mass of M\(_2\)CO\(_3\) is 1 g, so: \[ \text{Molar mass of M}_2\text{CO}_3 = \frac{\text{Mass}}{\text{Moles}} = \frac{1}{0.01} = 100 \, \text{g mol}^{-1}. \]
Final Answer: \( \boxed{100} \, \text{g mol}^{-1} \).
At STP \(x\) g of a metal hydrogen carbonate (MHCO$_3$) (molar mass \(84 \, {g/mol}\)) on heating gives CO$_2$, which can completely react with \(0.02 \, {moles}\) of MOH (molar mass \(40 \, {g/mol}\)) to give MHCO$_3$. The value of \(x\) is:
The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are
A. $ Cr^{2+} $
B. $ Fe^{2+} $
C. $ Fe^{3+} $
D. $ Co^{2+} $
E. $ Mn^{2+} $
Choose the correct answer from the options given below
Which of the following circuits has the same output as that of the given circuit?

Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).