NEET AIPMT SOLVED PAPER SCREENING 2012

  • question_answer
    50 mL of each gas A and of gas B takes 150 and 200 s respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gas A will be

    A) 96         

    B)                        128                        

    C)        32                          

    D)        64 No option is correct

    Correct Answer: D

    Solution :

    Given,                   \[{{V}_{A}}={{V}_{B}}=50\,mL\] \[{{T}_{A}}=150s\] \[{{T}_{B}}=200s\] \[{{M}_{B}}=36\] \[{{M}_{A}}=?\] From Graham's law of effusion            \[\frac{{{r}_{B}}}{{{r}_{A}}}=\sqrt{\frac{{{M}_{A}}}{{{M}_{B}}}}=\frac{{{V}_{B}}{{T}_{A}}}{{{T}_{B}}\cdot {{V}_{A}}}\] \[\Rightarrow \]               \[\sqrt{\frac{{{M}_{A}}}{36}}=\frac{{{V}_{A}}\times 150}{200\times {{V}_{A}}}\] or            \[\sqrt{\frac{{{M}_{A}}}{36}}=\frac{15}{20}=\frac{3}{4}\]                 \[\frac{{{M}_{A}}}{36}=\frac{9}{16}\] \[{{M}_{A}}=\frac{9\times 36}{16}=\frac{9\times 9}{4}=\frac{81}{4}=20.2\]  (No option is correct) Note If \[{{T}_{A}}=200s\]and\[{{T}_{B}}=150s,\] the \[{{M}_{A}}=64\]


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