JEE Main & Advanced Chemistry States of Matter Question Bank Ideal gas equation and Related gas laws

  • question_answer Volume of 4.4 g of \[C{{O}_{2}}\] at NTP is           [Pb. CET 1997]

    A)                 22.4 L    

    B)                 44.8 L

    C)                 2.24 L    

    D)                 4.48 L

    Correct Answer: C

    Solution :

               M.wt of \[C{{O}_{2}}\]  =  12+16+16 = 44                    Volume of 44 gm of \[C{{O}_{2}}\] at NTP = 22.4 litre                    1 gm of \[C{{O}_{2}}\] at NTP = \[\frac{22.4}{44}\]                    \[4.4\,gm\] of \[C{{O}_{2}}\] at N.T.P                                 \[\Rightarrow \frac{22.4}{44}\times 4.4\]litre \[=2.24\] litre


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