Two gases A and B are released from the two e | vedclass

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Question

(D) According to Graham's Law of Diffusion, the rate of diffusion is proportional to the distance traveled in the same time interval.$\frac{r_A}{r_B}

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) According to Graham's Law of Diffusion, the rate of diffusion is proportional to the distance traveled in the same time interval.$\frac{r_A}{r_B} = \frac{d_A}{d_B} = \sqrt{\frac{M_B}{M_A}}$Given: $d_A = 40 \ m$, $d_B = 100 - 40 = 60 \ m$, $M_B = 18$.Substituting the values:$\frac{40}{60} = \sqrt{\frac{18}{M_A}}$$\frac{2}{3} = \sqrt{\frac{18}{M_A}}$Squaring both sides:$\frac{4}{9} = \frac{18}{M_A}$$M_A = \frac{18 	imes 9}{4} = 40.5 \ g/mol$.Thus, the molecular weight of $A$ is $40.5 \ g/mol$.

Two gases $A$ and $B$ are released from the two ends of a tunnel of length $100 \ m$. $A$ travels $40 \ m$ before meeting the gas $B$ in the tunnel. If the molecular weight of $B$ is $18$, what is the molecular weight of $A$ (in $.5$)?

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