Two flasks A and B have equal volumes. A is maintained at 300   K and B at 600   K . Equal masses…

We know that the average kinetic energy depends on the temperature and number of moles. 

Now,  $\mathrm{K}.\mathrm{E}.$ expression for flasks$\mathrm{A}$ and $\mathrm{B}$ is given below:

${\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{A}}=\frac{3{\mathrm{n}}_{\mathrm{A}}{\mathrm{RT}}_{ \mathrm{A}}}{2}$

${\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{B}}=\frac{3{\mathrm{n}}_{\mathrm{B}}{\mathrm{RT}}_{\mathrm{B}}}{2}$
Assume common mass$=\mathrm{m}$
$$\begin{gathered} {\mathrm{n}}_{\mathrm{A}}=\mathrm{m}/2 \\ {\mathrm{n}}_{\mathrm{B}}=\mathrm{m}/44 \end{gathered}$$

$\frac{{\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{A}}}{{\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{B}}}=\frac{{\mathrm{n}}_{\mathrm{A}}{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{B}}{\mathrm{T}}_{\mathrm{B}}}=\frac{\mathrm{m}/2\times 300}{\mathrm{m}/44\times 600}=11:1$