1 g of a carbonate ( M2 CO3) on treatment with excess HCl produces 0.01 mol of CO2. The molar mass of M2 CO3 is

Given: Mass of M2 CO3 is 1 g and moles of CO2 produced are 0.01 mol. Find: The molar mass of M2 CO3. The reaction is M2 CO3 + 2 HCl 2 MCl + H2 O + CO2 From the stoichiometry, 1 mole of M2 CO3 produces 1 mole of CO2. Therefore, moles of M2 CO3 = 0.01 mol Now, Molar mass = mass moles = 10.01 = 100 g mol^-1 Therefore, the molar mass is 100 g mol^-1 and the correct option is A.

1 g of a carbonate ( M2 CO3) on treatment with excess HCl produce… — Solution | JEE Chemistry

MCQEasyJEE 2023Mole Concept

$1 \, \text{g}$ of a carbonate $\left(\text{M}_2\text{CO}_3\right)$ on treatment with excess HCl produces $0.01 \, \text{mol}$ of $\text{CO}_2$ . The molar mass of $\text{M}_2\text{CO}_3$ is _____

A
$100 \, \text{g/mol}$
B
$200 \, \text{g/mol}$
C
$250 \, \text{g/mol}$
D
$300 \, \text{g/mol}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Mass of $\text{M}_2\text{CO}_3$ is $1 \, \text{g}$ and moles of $\text{CO}_2$ produced are $0.01 \, \text{mol}$ .

Find: The molar mass of $\text{M}_2\text{CO}_3$ .

The reaction is

\text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{H}_2\text{O} + \text{CO}_2

From the stoichiometry, $1$ mole of $\text{M}_2\text{CO}_3$ produces $1$ mole of $\text{CO}_2$ .

Therefore,

\text{moles of } \text{M}_2\text{CO}_3 = 0.01 \, \text{mol}

Now,

\text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{1}{0.01} = 100 \, \text{g mol}^{-1}

Therefore, the molar mass is $100 \, \text{g mol}^{-1}$ and the correct option is A.

Common mistakes

Assuming the mole ratio between $\text{M}_2\text{CO}_3$ and $\text{CO}_2$ is not $1:1$ . This is wrong because one carbonate ion gives one molecule of $\text{CO}_2$ . Use the balanced equation before calculating moles.
Using the formula $\text{molar mass} = \frac{\text{moles}}{\text{mass}}$ instead of $\frac{\text{mass}}{\text{moles}}$ . This inverts the result. Always write the unit check to get $\text{g mol}^{-1}$ .

Practice more Mole Concept questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Mole Concept

Answer

Step-by-step solution

Common mistakes

Practice more Mole Concept questions

Related questions