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  4. 180 mL of hydrocarbon having the molecular weight 16 diffuses in 1.5 minutes. Under the same conditions, the time taken by 120 mL of sulphur dioxide to diffuse is :

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Q. 180 mL of hydrocarbon having the molecular weight 16 diffuses in 1.5 minutes. Under the same conditions, the time taken by 120 mL of sulphur dioxide to diffuse is :

EAMCETEAMCET 1996

Solution:

$ \frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}} $ or $ \frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}} $ or $ \frac{{{V}_{1}}}{{{t}_{1}}}\times \frac{{{t}_{2}}}{{{V}_{2}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}} $ or $ \frac{{{V}_{Hydrocarbon}}}{{{t}_{Hydrocarbon}}}\times \frac{{{t}_{S{{O}_{2}}}}}{{{V}_{S{{O}_{2}}}}}=\sqrt{\frac{{{M}_{S{{O}_{2}}}}}{{{M}_{Hydrocarbon}}}} $ $ {{M}_{S{{O}_{2}}}}=32+32=64 $ $ \frac{180}{1.5}\times \frac{{{t}_{S{{O}_{2}}}}}{120}=\sqrt{\frac{64}{16}} $ $ \frac{180\times {{t}_{S{{O}_{2}}}}}{1.5\times 120}=2 $ $ {{t}_{S{{O}_{2}}}}=\frac{2\times 1.5\times 120}{180}=2\,\min . $