J & K CET Medical J & K - CET Medical Solved Paper-2004

  • question_answer
    Number of atoms of oxygen present in 10.6 g of\[N{{a}_{2}}C{{O}_{3}}\]will be:

    A) \[6.02\times {{10}^{23}}\]                           

    B) \[12.04\times {{10}^{22}}\]

    C)  \[1.806\times {{10}^{23}}\]        

    D) \[31.80\times {{10}^{28}}\]

    Correct Answer: C

    Solution :

                    Molecular mass of \[N{{a}_{2}}C{{O}_{3}}=2\times 23+12+3\times 16=106\] \[\because \]\[106\text{ }g\text{ }N{{a}_{2}}C{{O}_{3}}\]contains \[=3\times 6.023\times {{10}^{23}}\]oxygen atoms \[\therefore \]10.6 g of\[N{{a}_{2}}C{{O}_{3}}\]will contain                 \[=\frac{3\times 6.023\times {{10}^{23}}}{106}\times 10.6\] \[=18.069\times {{10}^{24}}\] \[=1.806\times {{10}^{23}}\]oxygen atoms


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