Question

100 ml of O2 gas diffuse in 10 sec . 100 ml of gas X diffuse in T sec . gas X and time T can be

Answer 1

By Graham's law of diffusion:

Rate of Oxygen / Rate of X = √(MMX/MMO₂)

MMO₂ = 32

Rate of oxygen = 100/10 = 10

Rate of X = 100/T

10 × T/100 = 10T/100

1/10T = √(MMX/32)

1/100T² = MMX/32

MMX = 32/100T²

This is one equation with 2 unknowns. That is T and MMX.

It's a diophantine equation with several solutions:

These solutions could be.

T.......MMX

5.......8

15.....72

20....128

25....200

30....288

Answer 2

Answer:2

Explanation: