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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A gaseous mixture contains 56 g of \[{{\text{N}}_{\text{2}}}\text{, 44 g}\] of \[C{{O}_{2}}\]and 16 g of \[C{{H}_{4}}.\]The total pressure of mixture is 720 mm of Hg. The partial pressure of methane is :

    A)  75 aim                                 

    B)  160 atm

    C)  180 atm                              

    D)  215 atm

    Correct Answer: C

    Solution :

                     First of all we have to calculate the number of moles. Number of moles of \[{{N}_{2}}=\frac{56}{28}=2\] Number of moles of \[C{{O}_{2}}=\frac{44}{44}=1\] Number of moles of \[C{{H}_{4}}=\frac{16}{16}=1\] \[\therefore \]Total number of moles \[=2+1+1=4\] \[\therefore \] mole fraction of   \[C{{H}_{4}}=\frac{1}{4}\] \[\therefore \] partial pressure of \[C{{H}_{4}}\] = mole fraction of \[C{{H}_{4}}\times \]total pressure                 \[=\frac{1}{4}\times 720=180\,\text{atm}\]        


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