Answer
$M_{m}=10.60$ kg/kmol
Work Step by Step
We consider $100$ kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each
component are$$
\begin{aligned}
& m_{\mathrm{H} 2}=N_{\mathrm{H} 2} M_{\mathrm{H} 2}=(30 \mathrm{kmol})(2 \mathrm{~kg} / \mathrm{kmol})=60 \mathrm{~kg} \\
& m_{\mathrm{He}}=N_{\mathrm{He}} M_{\mathrm{He}}=(40 \mathrm{kmol})(4 \mathrm{~kg} / \mathrm{kmol})=160 \mathrm{~kg} \\
& m_{\mathrm{N} 2}=N_{\mathrm{N} 2} M_{\mathrm{N} 2}=(30 \mathrm{kmol})(28 \mathrm{~kg} / \mathrm{kmol})=840 \mathrm{~kg}
\end{aligned}
$$ The total mass is $$
m_m=m_{\mathrm{H} 2}+m_{\mathrm{He}}+N_{\mathrm{N} 2}=60+160+840=1060 \mathrm{~kg}
$$ Then the mass fractions are $$
\begin{aligned}
& \mathrm{mf}_{\mathrm{H} 2}=\frac{m_{\mathrm{H} 2}}{m_m}=\frac{60 \mathrm{~kg}}{1060 \mathrm{~kg}}=\mathbf{0 . 0 5 6 6 0} \\
& \mathrm{mf}_{\mathrm{He}}=\frac{m_{\mathrm{Hc}}}{m_m}=\frac{160 \mathrm{~kg}}{1060 \mathrm{~kg}}=\mathbf{0 . 1 5 0 9} \\
& \mathrm{mf}_{\mathrm{N} 2}=\frac{m_{\mathrm{N} 2}}{m_m}=\frac{840 \mathrm{~kg}}{1060 \mathrm{~kg}}=\mathbf{0 . 7 9 2 5}
\end{aligned}
$$ The apparent molecular weight of the mixture is $$
M_m=\frac{m_m}{N_m}=\frac{1060 \mathrm{~kg}}{100 \mathrm{kmol}}=10.60 \mathrm{~kg} / \mathbf{k m o l}
$$
