JEE Main & Advanced Chemistry Some Basic Concepts of Chemistry Question Bank Atomic Molecular and Equivalent masses

  • question_answer On reduction with hydrogen, 3.6 g of an oxide of metal left 3.2 g of metal. If the vapour density of metal is 32, the simplest formula of the oxide would be [DPMT 2004]

    A)                 \[MO\]    

    B)                 \[{{M}_{2}}{{O}_{3}}\]

    C)                 \[{{M}_{2}}O\]    

    D)                 \[{{M}_{2}}{{O}_{5}}\]

    Correct Answer: C

    Solution :

               As  we know that                    Equivalent weight \[=\frac{\text{weight of metal}}{\text{weight of oxygen}}\times 8\]                                                \[=\frac{32}{0.4}\times 8=64\]                    Vapour density \[=\frac{\text{mol}\text{. wt}}{\text{2}}\]                    Mol. wt \[=2\times V.D=2\times 32=64\]                    As we know that \[n=\frac{\text{mol}\text{.}\,\text{wt}}{\text{eq}\text{. wt}}=\frac{64}{64}=1\]                                 Suppose, the formula of metal oxide be \[{{M}_{2}}{{O}_{n}}\]. Hence the formula of metal oxide \[={{M}_{2}}O\].


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