Answer
$M_{w}=28.99$ kg/kmol
Work Step by Step
Standard air is taken as $79 \%$ nitrogen and $21 \%$ oxygen by mole. That is,
$$
\begin{aligned}
& y_{\mathrm{O} 2}=0.21 \\
& y_{\mathrm{N} 2}=0.79
\end{aligned}
$$ Adding another $0.05$ moles of $\mathrm{O}_2$ to $1 \mathrm{kmol}$ of standard air gives
$$
\begin{aligned}
& y_{\mathrm{O} 2}=\frac{0.26}{1.05}=0.2476 \\
& y_{\mathrm{N} 2}=\frac{0.79}{1.05}=0.7524
\end{aligned}
$$ Then,
$$
M_{\text {w }}=y_{\mathrm{O} 2} M_{\mathrm{C} 2}+y_{\mathrm{N} 2} M_{\mathrm{N} 2}=0.2476 \times 32+0.7524 \times 28=\mathbf{2 8 . 9 9} \mathbf{k g} / \mathbf{k m o l}
$$
