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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    1 gram of a carbonate \[({{M}_{2}}C{{O}_{3}})\] on treatment with excess HCl produces 0.01186 mole of \[C{{O}_{2}}.\]the molar mass of \[{{\text{M}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\]in \[\text{g}\,\text{mo}{{\text{l}}^{-1}}\]is:-                               JEE Main Solved Paper-2017

    A)  1186                     

    B)  84.3

    C)  118.6                   

    D)  11.86

    Correct Answer: B

    Solution :

     Given chemical \[\text{e}{{\text{q}}^{\text{n}}}\] \[\underset{1\,gm}{\mathop{{{M}_{2}}C{{O}_{3}}}}\,+2HCl\xrightarrow{{}}2MCl+{{H}_{2}}O+\underset{0.01186\,\text{mol}}{\mathop{C{{O}_{2}}}}\,\]\[\Rightarrow \]from the balanced chemical \[e{{q}^{n}}.\] \[\frac{1}{M}=0.01186\]                 \[\Rightarrow \]\[\,gm/mol\]


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