180ml of a hydrocarbon having the molecular weight 16 diffuses in 1.5 min. Under similar conditions time taken by 120 ml of $S O_{2}$ to diffuse is ______________ sec.

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Detailed Solution

According to Graham's law,

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$ , where $r_{1}$ and $r_{2}$ are the rates of diffusion of gases 1 and 2 respectively.

and, $M_{1}$ and $M_{2}$ are the molecular weights of gases 1 and 2 respectively

Rate=time required for diffusion volume of gas

$r_{1} = \frac{180}{1.5} = 120 r_{2} = \frac{t_{2}}{120} t_{2} = \frac{120}{60} = 2$

Therefore, the required time is 2 minute = 120 sec.

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