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JEE Main 20172 April
Exam
JEE Main
Year
2017
Date
2 April
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2017 · 2 April · Chemistry · Some Basic Concepts of Chemistry

1 gram of a carbonate ( M _ 2 CO _ 3 ) on treatment with excess HCl...

11 gram of a carbonate (M2CO3)\left({M}_{2}{CO}_{3}\right) on treatment with excess HClHCl produces 0.011860.01186 moles of CO2{CO}_{2} . The molar mass of M2CO3ingmol−1{M}_{2}{CO}_{3} in {gmol}^{- 1} is:

Options

  1. A84.3
  2. B118.6
  3. C11.86
  4. D1186

Correct answer

A. 84.3

Step-by-step solution

{M}_{2}C{O}_{3} + 2HCl \to 2MCl + {H}_{2}O + C{O}_{2}<br/><br/>Fromreaction,<br/><br/>MolesofM2CO3=MolesofCO2<br /><br />From reaction,<br /><br />Moles of{M}_{2}C{O}_{3} = MolesofC{O}_{2}

WeightofM2CO3MolarMassofM2CO3=WeightofCO2MolarmassofCO2\dfrac{Weightof{M}_{2}{CO}_{3}}{MolarMassof{M}_{2}{CO}_{3}} = \dfrac{Weightof{CO}_{2}}{Molarmassof{CO}_{2}}

1MolarMassofM2CO3=0.01186×4444\dfrac{1}{MolarMassof{M}_{2}{CO}_{3}} = \dfrac{0 . 01186 \times 44}{44}

1MolarMassofM2CO3=0.01186\dfrac{1}{MolarMassof{M}_{2}{CO}_{3}} = 0 .01186

Molar Mass =10.01186=84.3= \dfrac{1}{0.01186} = 84.3 g/mol

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