Under identical conditions of temperature and pressure the ratio of the rates of effusion of $O_{2}$ and $C O_{2}$ gases is given by?

\frac{r a t e o f e f f i s i o n o f o x y g e n}{r a t e o f e f f i s i o n o f C O_{2}} = 0.87

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\frac{r a t e o f e f f i s i o n o f o x y g e n}{r a t e o f e f f i s i o n o f C O_{2}} = 1.17

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\frac{r a t e o f e f f i s i o n o f o x y g e n}{r a t e o f e f f i s i o n o f C O_{2}} = 8.7

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\frac{r a t e o f e f f i s i o n o f o x y g e n}{r a t e o f e f f i s i o n o f C O_{2}} = 0.117

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Solution

The correct option is A $\frac{r a t e o f e f f i s i o n o f o x y g e n}{r a t e o f e f f i s i o n o f C O_{2}} = 0.87$
Given,
Temperature and pressure are constant
Molar mass of oxygen $= (2 \times 16) g / m o l e$
Molar mass of carbon dioxide $= 44 g / m o l e$
Now,
According to graham's law of diffusion
Rate of diffusion is inversely proportional to the square root of molar mass
$r \propto \frac{1}{\sqrt{M}}$
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{44}{32}} = \sqrt{1.375} = 1.17$

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