Answer :

Answer

Molar mass of the gas sample = 43.9968 g/mol

Explanation

The ideal gas law equation is given by

[tex]\begin{gathered} PV=nRT \\ \end{gathered}[/tex]

Where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the molar gas constant.

At STP, T = 273 K and P = 1 atm

Note that the number of moles, n can be rewritten as

[tex]\begin{gathered} n=\frac{m}{M} \\ \\ where\text{ }m\text{ }is\text{ }the\text{ }mass\text{ }and\text{ }M\text{ }is\text{ }the\text{ }molar\text{ }mass \end{gathered}[/tex]

Therefore, the ideal gas law equation above becomes

[tex]\begin{gathered} PV=\frac{m}{M}RT \\ \\ \Rightarrow M=\frac{mRT}{PV} \end{gathered}[/tex]

Putting the values of the given parameters below into the formula, we have

[tex]\begin{gathered} P=1\text{ }atm \\ \\ V=50\text{ }L \\ \\ T=273\text{ }K \\ \\ m=98.2g \\ \\ R=0.082057338\text{ }L.atm.\text{/}K.mol \end{gathered}[/tex][tex]\begin{gathered} M=\frac{98.2\text{ }g\times0.082057338\text{ }L.atm\text{/}K.mol\times273\text{ }K}{1\text{ }atm\times50\text{ }L} \\ \\ M=43.9968\text{ }g\text{/}mol \end{gathered}[/tex]

Therefore the molar mass of 98.2 g sample of gas that fills a 50 liter container at STP is 43.9968 g/mol

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Rewritten by : Barada