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EAMCET Medical EAMCET Medical Solved Paper-1998

  • question_answer
    The volume of 1.8 g of carbon monoxide at \[\text{27}{{\,}^{\text{o}}}\text{C}\,\]and 0.821 atm pressure is \[\text{(R = 0}\text{.0821}\,L\,atm\,{{K}^{-1}}\,\text{mo}{{\text{l}}^{-1}}\text{)}\]

    A)   1.5L                                     

    B)  0.3L

    C)  3L                                          

    D)  30L

    Correct Answer: C

    Solution :

                     We know that \[PV=nRT\]                 Number of moles of \[C{{O}_{2}}\] \[(n)=\frac{\text{Wt}\text{.of}\,C{{O}_{2}}}{\text{mol}\text{.Wt}\text{.of}\,\text{C}{{\text{O}}_{\text{2}}}}\] \[=\frac{2.8}{28}=0.1\] \[T=27+273=300K\] \[P=0.821\,atm\] \[R=0.0821\]                 Then, \[V=\frac{nRT}{P}=\frac{0.1\times 0.0821\times 300}{0.821}\]                                 \[=3\,L\]


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