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NEET Sample Paper NEET Sample Test Paper-87

  • question_answer
    \[A+2B+3C\rightleftharpoons A{{B}_{2}}{{C}_{3}}\] Reaction of 6.0 g of A, \[6.0\times {{10}^{23}}\] atoms of B, and 0.036 mol ofC yields 4.8 g of compound \[A{{B}_{2}}{{C}_{3}}.\] If the atomic mass of A and C are 60 and 80 amu, respectively, the atomic mass of B is (Avogadro no. \[=6\times {{10}^{23}}mo{{l}^{-1}}\]):

    A) 50 amu

    B)        60 amu

    C) 70 amu

    D)        40 amu

    Correct Answer: A

    Solution :

    [a] \[A+3B+3CA{{B}_{2}}{{C}_{3}}\]??(l) No. of moles of A\[=\frac{6.0\,g}{60\,g/mol}=0.1\,mol\] No. of moles of B\[=\frac{6.00\times {{10}^{23}}}{6.000\times {{10}^{23}}}=1\,mol\] No. of moles of C \[=0.036\] \[A{{B}_{2}}{{C}_{3}}\] formed accordingly to C which is a limiting reagent. Since 3 moles of C are used in (1) So it gives 1 mole of \[A{{B}_{2}}{{C}_{3}}\] \[{{n}_{A{{B}_{2}}{{C}_{3}}}}=\frac{0.036}{3}=0.012\] \[\text{=}\frac{\text{Given}\,\,\text{mass(4}\text{.8)}}{\text{Molecular}\,\,\text{mass(M}\text{.M)}}\] Mol. mass \[=\frac{4.8}{0.012}=400\] \[\Rightarrow \]\[400=60+(2\times x)+(80\times 3)\] \[\Rightarrow \]\[x=50\]


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