WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    The total number of protons in \[10g\] of calcium carbonate is \[({{N}_{0}}=6.023\times {{10}^{23}}):\]

    A)  \[3.01\times {{10}^{24}}\]          

    B)  \[~4.06\times {{10}^{24}}\]

    C)  \[2.01\times {{10}^{24}}\]          

    D)  \[3.02\times {{10}^{24}}\]

    Correct Answer: A

    Solution :

                     We know that protons in 1 mole \[CaC{{O}_{3}}\] = atomic number of calcium + atomic number of carbon + 3 (atomic number of oxygen) \[=20+6+3(8)=50\,mol\]                 \[\therefore \] proton in 10 g \[CaC{{O}_{3}}=\frac{10\times 50}{100}\times 6.02\times {{10}^{23}}\]                 \[=3.01\times {{10}^{24}}\]


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