EAMCET Medical EAMCET Medical Solved Paper-2008

  • question_answer
    In a flask of V litres, 0.2 moles of \[{{\text{O}}_{\text{2}}}\text{,0}\text{.4}\]moles of \[{{\text{N}}_{2}},\,0.1\]moles of \[\text{N}{{\text{H}}_{\text{3}}}\] and 0.3 moles of He gases are present at \[\text{27}{{\,}^{\text{o}}}\text{C}\text{.}\]If total pressure exerted by these non-reacting gases is 1 atm, the partial pressure exerted by \[{{\text{N}}_{2}}\]gas is

    A)  0.4 atm                               

    B)  0.3 atm

    C)  0.2 atm                               

    D)  0.1 atm

    Correct Answer: A

    Solution :

                     Given,                                        Total pressure = 1 atm Moles of oxygen =0.2 Moles of nitrogen =0.4 Moles of \[N{{H}_{3}}=0.1\] Moles of He =0.3 Total number of moles = 0.2 + 0.4 + 0.1 4- 0.3 =1.0 Partial pressure of nitrogen                     = total pressure \[\times \] mole fraction of nitrogen moles of nitrogen \[\text{= 1}\,\text{atm  }\!\!\times\!\!\text{  }\frac{\text{moles of nitrogen}}{\text{total number of moles}}\] \[=1\times \frac{0.4}{1}=0.4\,\,\text{atm}\]


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