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JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. \[20d{{m}^{3}}\]of \[S{{O}_{2}}\] diffuses through the porous 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):     JEE Main Online Paper (Held On 19 April 2016)

    A) 7.09                       

    B) 14.1

    C) 10.0                                       

    D) 28.2

    Correct Answer: B

    Solution :

    According to Graham?s Law Diffusion: \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] Since rate of diffusion = \[\text{=}\frac{\text{Vol}\text{. of gas diffused }\left( \text{V} \right)}{\text{Time}\,\text{taken}\,\text{for}\,\text{diffusion}\,\text{(t)}}\] \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}\]or \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] \[=\frac{20/60}{{{V}_{2}}/30}=\sqrt{\frac{16/2}{32/2}}=\sqrt{\frac{1}{2}}\] \[\because \]Mol. \[wt=2\times V.D\] \[\therefore \]\[V.D.=\frac{Mol.wt}{2}\] On calculating, \[{{V}_{2}}=14.1\]


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