As the kinetic energy expression used the $\mathrm{RMS}$ speed of the gas as:$\begin{array}{ccc}\frac{1}{2}{{\mathrm{Mv}}^2}_{\mathrm{rms}}& =& 5.0\times {10}^4\\ \frac{1}{2}\mathrm{M}(5.0\times {10}^4\times {10}^{-2}{{)}^2}_{\mathrm{rms}}& =& 4\times {10}^3\\ \mathrm{M}& =& 0.032 \mathrm{Kg} {\mathrm{mole}}^{-1}\end{array}$

Therefore, molecular weight is $0.032 \mathrm{Kg} \mathrm{per} \mathrm{mole} \mathrm{or} 32 \mathrm{g} \mathrm{per} \mathrm{mole}$.