$H X_{2}$ (2) and $O X_{2}$ (32) are released simultaneously at the two ends of a $100 cm$ tube. Assuming the gases travel at rates given by Graham's law, the distance in cm (nearest integer) from the $H X_{2}$ end at which they meet is

Numerical answer type, answer in cm, tolerance 0.5.

Answer: 80 cm

r_{H X_{2}} / r_{O X_{2}} = 32/2 = 4

. Distances covered in the same time are in the ratio

4 : 1

. From the

H X_{2}

end:

100 \times \frac{4}{5} = 80 cm