Kinetic energy of translation =23nRT $\begin{aligned}
& n=\frac{50 \mathrm{g}}{44 \mathrm{g}}=\frac{25}{22} \mathrm{mol} \
& T=17^{\circ} \mathrm{C}=290 \mathrm{K}
\end{aligned}\RightarrowKineticenergyoftranslation\begin{aligned}
& =\frac{3}{2}\left(\frac{25}{22}\right)(8.3)(290) \mathrm{J} \
& =4102.8 \mathrm{~J}
\end{aligned}$