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(A) According to Graham's Law of Diffusion,for gases diffusing in the same time $t$ under identical conditions of temperature and pressure,the rate of

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$100$

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(A) According to Graham's Law of Diffusion,for gases diffusing in the same time $t$ under identical conditions of temperature and pressure,the rate of diffusion $r$ is proportional to the volume $V$ diffused $(r = V/t)$.Thus,$\frac{r_A}{r_B} = \frac{V_A}{V_B} = \sqrt{\frac{M_B}{M_A}}$.Given $V_A = 50 \ mL$,$V_B = 40 \ mL$,and $M_A = 64$.Substituting the values: $\frac{50}{40} = \sqrt{\frac{M_B}{64}}$.Squaring both sides: $(\frac{5}{4})^2 = \frac{M_B}{64}$.$\frac{25}{16} = \frac{M_B}{64}$.$M_B = \frac{25}{16} 	imes 64 = 25 	imes 4 = 100$.

Under similar conditions of temperature and pressure,$50 \ mL$ of gas $A$ diffuses through a hole in the same time that $40 \ mL$ of gas $B$ diffuses. If the molar mass of gas $A$ is $64$,what is the molar mass of gas $B$?

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