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NEET Chemistry States of Matter / पदार्थ की अवस्थाएँ - गैस एवं द्रव Question Bank Self Evaluation Test - States of Matter

  • question_answer
    Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. \[20d{{m}^{3}}\] of \[S{{O}_{2}}\] diffuses through the Dorous partition in 60 seconds. The volume of \[{{O}_{2}}\] in \[d{{m}^{3}}\] which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32 u):

    A) 7.09                 

    B) 14.1  

    C) 10.0                 

    D) 28.2

    Correct Answer: B

    Solution :

    [b] According to Graham's Law Diffusion: \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] Since rate of diffusion \[=\frac{Vol.\text{ }of\text{ }gas\text{ }diffused\left( V \right)}{Time\,taken\,for\,diffusion(t)}\] \[\therefore \frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}\] Or \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] \[=\frac{20/60}{{{V}_{2}}/30}=\sqrt{\frac{16/2}{32/2}}=\sqrt{\frac{1}{2}}\] \[\therefore \] Mol. \[wt=2\times V.D\] \[\therefore V.D=\frac{Mol.wt}{2}\] On calculating, \[{{V}_{2}}=14.1\]


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