3.2 , g of oxygen gas diffuses in 10 , minute | vedclass

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(B) According to Graham's Law of Diffusion,the rate of diffusion $r \propto \frac{W}{t \sqrt{M}}$.Therefore,$\frac{r_{O_{2}}}{r_{N_{2}}} = \frac{W_{O_

Vedclass · Accepted Answer

$8.2$

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(B) According to Graham's Law of Diffusion,the rate of diffusion $r \propto \frac{W}{t \sqrt{M}}$.Therefore,$\frac{r_{O_{2}}}{r_{N_{2}}} = \frac{W_{O_{2}} / t_{O_{2}}}{W_{N_{2}} / t_{N_{2}}} = \sqrt{\frac{M_{N_{2}}}{M_{O_{2}}}}$Given: $W_{O_{2}} = 3.2 \, g$,$t_{O_{2}} = 10 \, minute$,$M_{O_{2}} = 32 \, g/mol$,$W_{N_{2}} = 2.8 \, g$,$M_{N_{2}} = 28 \, g/mol$.Substituting the values: $\frac{3.2 / 10}{2.8 / t_{N_{2}}} = \sqrt{\frac{28}{32}}$$\frac{3.2 	imes t_{N_{2}}}{28} = \sqrt{0.875} \approx 0.9354$$t_{N_{2}} = \frac{0.9354 	imes 28}{3.2} \approx 8.185 \, minute \approx 8.2 \, minute$.

$3.2 \, g$ of oxygen gas diffuses in $10 \, minute$. Under similar conditions,$2.8 \, g$ of nitrogen gas will diffuse in ............ $minute$.

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