JEE Main & Advanced Chemistry Some Basic Concepts of Chemistry Question Bank Atomic Molecular and Equivalent masses

  • question_answer Ratio of \[{{C}_{p}}\]and \[{{C}_{v}}\]of a gas X is 1.4, the number of atom of the gas ?X? present in 11.2 litres of it at NTP will be [CBSE 1999]

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

    B)                 \[1.2\times {{10}^{23}}\]

    C)                 \[3.01\times {{10}^{23}}\]              

    D)                 \[2.01\times {{10}^{23}}\]

    Correct Answer: A

    Solution :

               \[\frac{{{C}_{P}}}{{{C}_{V}}}=1.4\] so, given gas is diatomic                    11.2L \[=3.01\times {{10}^{23}}\]molecules                                 \[\therefore \] No. of atoms \[=3.01\times {{10}^{23}}\times 2\] \[=6.023\times {{10}^{23}}\]atoms

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