Equal volumes of $H X_{2}$ and $O X_{2}$ are placed at the two ends of a tube of length $90 cm$ at the same instant. If the gases diffuse toward each other with rates given by Graham's law, they first meet at a distance from the $H X_{2}$ end of

Answer: A

\frac{r _{H X_{2}}}{r _{O X_{2}}} = 32/2 = 4

. In the same time the distances are in the ratio

4 : 1

. From the

H X_{2}

end:

90 \times \frac{4}{5} = 72 cm