Equal volume of 2 gases A and B diffuse in 20 | vedclass

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(A) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M_w$: $\frac{r_A}

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$20$

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(A) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M_w$: $\frac{r_A}{r_B} = \sqrt{\frac{M_{wB}}{M_{wA}}}$ Since rate $r = \frac{V}{t}$,and volumes are equal $(V_A = V_B)$: $\frac{t_B}{t_A} = \sqrt{\frac{M_{wB}}{M_{wA}}}$ Substituting the given values ($t_A = 20 \ s$,$t_B = 10 \ s$,$M_{wA} = 80$): $\frac{10}{20} = \sqrt{\frac{M_{wB}}{80}}$ $0.5 = \sqrt{\frac{M_{wB}}{80}}$ Squaring both sides: $0.25 = \frac{M_{wB}}{80}$ $M_{wB} = 0.25 	imes 80 = 20$

Equal volume of $2$ gases $A$ and $B$ diffuse in $20 \ s$ and $10 \ s$ respectively. If molar mass of $A$ is $80$,then the molar mass of $B$ is:

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