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

  • question_answer 1.24 gm P is present in 2.2 gm

    A)                 \[{{P}_{4}}{{S}_{3}}\]     

    B)                 \[{{P}_{2}}{{S}_{2}}\]

    C)                 \[P{{S}_{2}}\]     

    D)                 \[{{P}_{2}}{{S}_{4}}\]

    Correct Answer: A

    Solution :

               Choice A is \[{{P}_{4}}{{S}_{3}}\]                    \[\because \]   \[\frac{31\times 4}{(124)}gm\]P is present in 220gm \[{{P}_{4}}{{S}_{3}}\]                                 \[\therefore \]   1.24gm P is present in = \[\frac{220}{124}\times 1.24=2.2gm\]


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