EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
    A vessel contains equal volumes of \[\text{S}{{\text{O}}_{\text{2}}}\]and \[\text{C}{{\text{H}}_{4}}.\]Through a small hole the gases effused into vacuum. After 200 seconds the total volume is reduced to half. What is the ratio of \[\text{S}{{\text{O}}_{\text{2}}}\]and \[\text{C}{{\text{H}}_{\text{4}}}\]remaining in the vessel?

    A)  1:2                                        

    B)  2:1

    C)  1 :1                                       

    D)  1:4

    Correct Answer: B

    Solution :

                     According to Grahams law of diffusion, \[\frac{{{r}_{S{{O}_{2}}}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{{{M}_{C{{H}_{4}}}}}{{{M}_{S{{O}_{2}}}}}}\]                 \[\frac{{{r}_{S{{O}_{2}}}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{16}{64}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\] \[{{r}_{S{{O}_{2}}}}:{{r}_{C{{H}_{4}}}}=1:2\] \[\text{S}{{\text{O}}_{\text{2}}}\] and \[\text{C}{{\text{H}}_{4}}\] effused in the ratio 1 : 2. So, the amount remains in the vessel will be in the ratio as \[S{{O}_{2}}:C{{H}_{4}}=2:1\]


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