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KVPY Sample Paper KVPY Stream-SX Model Paper-6

  • question_answer
    At \[27{}^\circ C\], hydrogen is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same temperature and pressure as that of I-L is leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atmosphere. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litres, the molecular weight of the unknown gas is:

    A) 516                              

    B) 2066

    C) 5033                            

    D) 1033

    Correct Answer: D

    Solution :

    using gas equation; \[PV=nRT\]
    Total no. of moles of gas in the mixture (n) \[=\frac{PV}{RT}=\frac{6\times 3}{0.0821\times 300}=0.7308mol.\]
    Thus no. of moles of unknown gas \[=0.7308-0.7=0.0308mol.\]
    Now we know that \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{moles\,of\,hydrogen\,gas}{moles\,of\,unknown\,gas}=\frac{0.7}{0.0308}\]
    Also we know that \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{\sqrt{{{M}_{2}}}}{\sqrt{{{M}_{1}}}}\]
    \[\therefore \]\[{{M}_{2}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\times {{M}_{1}}\]
    Or \[{{M}_{2}}={{\left( \frac{0.7}{0.0308} \right)}^{2}}\times 2=1033\]


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