Question

The rate of diffusion of methane at given temp is twice of a gas x the molecular wieght of gas x is

Answer 1

The rate of diffusion of methane at given temp is twice of a gas x the molecular weight of gas x is 64.

The rate of diffusion of two gases are compared as follows.

$\frac{r1}{r2}  = \sqrt{\frac{M1}{M2}}$

As The rate of diffusion of methane at given temp is twice of a gas x the molecular weight of gas x is 64.

Hence,

$\frac{r1}{2r1}  = \sqrt{\frac{16}{M2}}$

$\frac{1}{2}  = \sqrt{\frac{16}{M2}}$

Squaring both sides to remove the square root.

$(\frac{1}{2})^{2}   = (\sqrt{\frac{16}{M2}})^{2}$

Then we will get after squaring both sides,Then we will get after squaring both sides,

$\frac{1}{4}  = \frac{16}{M2}$

M2 = 64

Detailed solution and explanation is given in the attachment