# Question #4b1ce

##### 1 Answer

#### Answer:

#### Explanation:

As you know, the **density** of a substance, *one unit of volume* of said substance.

In your case, you know that an unknown gas has a density of **STP conditions**, which are defined as

#"pressure" = "100 kPa" = 100/101.325# #"atm"#

#"temperature" = 0^@"C" = "273.15 K"#

This tells you that **every**

#color(blue)(rho = m/V)#

where **mass** of the gas and **volume** it occupies, tells you the density of the gas.

At this point, your tool of choice will be the **ideal gas law equation**

#color(blue)(ul(color(black)(PV = nRT)))#

Here

#P# is the pressure of the gas#V# is the volume it occupies#n# is the number of moles of gas present in the sample#R# is theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

If you use the fact that the number of moles of gas can be expressed as the ratio between the **mass** of the sample and the **molar mass** of the gas

#n = m/M_M#

you can rewrite the ideal gas law equation as

#PV = m/M_M * RT#

Rearrange this to isolate

#M_M = color(blue)(m/V) * (RT)/P#

This is equivalent to

#M_M = color(blue)(rho) * (RT)/P#

Plug in the density of the gas and the aforementioned conditions for pressure and temperature to find the molar mass of the gas

#M_M = "1.429 g" color(red)(cancel(color(black)("L"^(-1)))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 273.15 color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#

#color(darkgreen)(ul(color(black)(M_M = "32.47 g mol"^(-1))))#

The answer is rounded to four **sig figs**, the number of sig figs you have for the density of the gas.

**SIDE NOTE** *A lot of sources still use the old definition of STP conditions*

#"pressure = 1 atm"# #"temperature" = 0^@"C" = "273.15 K"#

*so if this is the definition given to you, redo the calculations using* *instead of* *for the pressure of the gas.*