Answer:
The answer is "[tex]34 \ \frac{g}{mol}[/tex]"
Explanation:
Please find the complete question in the attached file.
Formula:
[tex]\to PV = nRT\\\\n = \frac{mass}{MW} \\\\PV = \frac{mRT}{MW}\\\\MW = \frac{mRT}{(PV)}\\\\MW = \frac{(0.941)(0.082)(150+273)}{(1 \times 0.96)}[/tex]
[tex]= \frac{(0.941)(0.082)(423)}{(0.96)} \\\\= \frac{32.639526}{(0.96)} \\\\= 33.9995062\ \ \ or \ \ \ 34 \ \frac{g}{mol}[/tex]
