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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    An oxide \[{{A}_{x}}{{O}_{y}}\] has molecular weight 288 u. Atomic weights of A and 0 respectively are 12 and 16. The formula of the compound, if A is 50% by weight is

    A) \[{{A}_{3}}{{O}_{5}}\]          

    B) \[{{A}_{5}}{{O}_{3}}\]

    C) \[{{A}_{12}}{{O}_{9}}\]         

    D) \[{{A}_{9}}{{O}_{12}}\]

    Correct Answer: C

    Solution :

    [c] : Molar ratio \[A:O\frac{50}{12}:\frac{50}{16}=4.16:3.12\] \[=1.33:1=4:3\] \[E.F.={{A}_{4}}{{O}_{3}}\] \[E.F.W.=\left( 12\times 4 \right)+\left( 16\times 3 \right)=96\] \[n=\frac{M.F.W}{E.F.W}=\frac{288}{96}=3\] Molecular formula \[={{({{A}_{4}}{{O}_{3}})}_{3}}=A{{ & }_{12}}{{O}_{9}}\]


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