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JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
                                                      An unknown chlorohydrocarbon has 3.55 % of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1g of chlorohydrocarbon are: (Atomic wt. of\[Cl=35.5u\]); Avogadro constant\[=6.023\times {{10}^{23}}mo{{l}^{-1}}\])[JEE Main 16-4-2018]

    A) \[6.023\times {{10}^{23}}\]            

    B) \[6.023\times {{10}^{21}}\]

    C) \[6.023\times {{10}^{9}}\]              

    D) \[6.023\times {{10}^{20}}\]

    Correct Answer: D

    Solution :

     An unknown chlorohydrocarbon has 3.55% of chlorine. 100 g of chlorohydrocarbon has 3.55 g of chlorine. g of chlorohydrocarbon will have \[3.55\times \frac{1}{100}=0.0355\,g\]of chlorine. Atomic wt. of\[Cl=35.5\,g/mol\] Number of moles of \[Cl=\frac{0.0355g}{35.5g/mol}=0.001\,mol\] Number of atoms of \[Cl=0.001\,mol\times 6.023\times {{10}^{23}}mo{{l}^{-1}}=6.023\times {{10}^{20}}\]


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