Question

a gas a having molecular weight 4 diffuser thrice as fast as the gas B the molecular weight of gas b is what​

Answer 1

Answer:

the molecular weight of the gas Bis12

Answer 2

Answer:

Answer is 36.

Explanation:

Graham’s law actually solve this equation. Graham’s law state that rate of distribution of gas is inverse proportional to square root of its molecular weight.

Molecular weight of A= 4

Let, rate of distribution of gas B = x

Then, rate of distribution of gas A=3x

Let, molecular weight of gas B = M

According to Graham’s Law:

$\frac{3 x}{x}=\sqrt{\frac{M}{4}}$

Or, $\frac{\sqrt{M}}{2}=3$

Or,  $\sqrt{M}=6$

Then, M =36

So, molecular weight of gas B is 36.