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Uttarakhand PMT Uttarakhand PMT Solved Paper-2005

  • question_answer
    The volume of 2.8 gm of carbon monoxide at \[27{}^\circ C\]and 0.821 atm pressure is [R = 0.0821 litre atm \[mo{{l}^{-1}}{{K}^{-1}}\]]:

    A)  0.5 litre            

    B)  1.51itrc

    C)  3 litre              

    D) 6 litre

    Correct Answer: C

    Solution :

     The volume of carbon monoxide\[=\frac{nRT}{P}\] Mass of carbon monoxide = 2.8 g \[P=0.821\text{ }atm,\text{ }T=27+273=300\text{ }K\] \[R=0.0821\] \[\because \]gram molecular weight of \[CO=12+16=28\] \[\therefore \]number of moles in 2.8 g of \[CO=\frac{2.8}{28}=0.1\] \[\therefore \]volume\[=\frac{0.1\times 0.0821\times 300}{0.821}=3\]litre


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