Answer
(a) $53.33\frac{lb}{lb/space mole}$
(b) $0.4548$, $0.5452$
Work Step by Step
(a) We can find the required molecular weight as follows:
$n=\frac{PV}{RT}$
$\implies n=\frac{40\times 45 \times 144}{1545.35(150+459.67)}$
This simplifies to:
$n=0.275lb\space mole$
Now $M=\frac{m}{n-\frac{m_{O_2}}{M_{O_2}}}$
We plug in the known values to obtain:
$M=\frac{8}{0.275-\frac{4}{32}}$
This simplifies to:
$M=53.33 \space \frac{lb}{lb\space mole}$
(b) The required mole fraction can be determined as follows:
First for oxygen
$y_{O_2}=\frac{\frac{4}{12}}{\frac{32}{0.0229}}$
This simplifies to:
$y_{O_2}=0.4548$
Now, for unknown gas
$y=\frac{{8}{12}}{\frac{53.33}{0.0229}}$
This simplifies to:
$y=0.5452$
