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Some Basic Concepts of Chemistry question

2017 · Shift 0 · Q24
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Some Basic Concepts of Chemistry question

2017 · Shift 0 · Q24

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
1 gram of a carbonate (M2CO3M_2CO_3M2​CO3​) on treatment with excess HClHClHCl produces 0.01186 mole of CO2CO_2CO2​. The molar mass of M2CO3M_2CO_3M2​CO3​ in g mol–1 is:
  1. A
    84.3
  2. B
    118.6
  3. C
    11.86
  4. D
    1186
View written solutionFree

Correct answer: A

  1. Write the reaction

For a carbonate of the form M2CO3M_2CO_3M2​CO3​ reacting with excess hydrochloric acid:

M2CO3+2HCl→2MCl+H2O+CO2M_2CO_3 + 2HCl \rightarrow 2MCl + H_2O + CO_2M2​CO3​+2HCl→2MCl+H2​O+CO2​

From the equation, 1 mole of M2CO3M_2CO_3M2​CO3​ gives 1 mole of CO2CO_2CO2​.

  1. Use the given moles of CO2CO_2CO2​

Given:

n(CO2)=0.01186 moln(CO_2)=0.01186\ \text{mol}n(CO2​)=0.01186 mol

So,

n(M2CO3)=0.01186 moln(M_2CO_3)=0.01186\ \text{mol}n(M2​CO3​)=0.01186 mol

  1. Calculate molar mass of M2CO3M_2CO_3M2​CO3​

Given mass of carbonate = 111 g.

Using

Molar mass=massmoles\text{Molar mass} = \frac{\text{mass}}{\text{moles}}Molar mass=molesmass​

Molar mass of M2CO3=10.01186\text{Molar mass of } M_2CO_3 = \frac{1}{0.01186}Molar mass of M2​CO3​=0.011861​

=84.3 g mol−1=84.3\ \text{g mol}^{-1}=84.3 g mol−1

  1. Match with options

84.3 g mol−184.3\ \text{g mol}^{-1}84.3 g mol−1

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They match.

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