Question

use ideal gas law to find the molar mass of 98.2 g sample of gas that fills a 50 liter container at stp

Answer 1

Molar mass of the gas sample = 43.9968 g/mol

Step-by-step explanation

The ideal gas law equation is given by

`\begin{gathered} PV=nRT \\ \end{gathered}`

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

`\begin{gathered} n=(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}`

Therefore, the ideal gas law equation above becomes

`\begin{gathered} PV=(m)/(M)RT \\ \\ \Rightarrow M=(mRT)/(PV) \end{gathered}`

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

`\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}`
`\begin{gathered} M=\frac{98.2\text{ }g*0.082057338\text{ }L.atm\text{/}K.mol*273\text{ }K}{1\text{ }atm*50\text{ }L} \\ \\ M=43.9968\text{ }g\text{/}mol \end{gathered}`

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