# A gas is found to have a density of #"1.80 g/L"# at #"76 cm Hg"# and #27^@ "C"#. The gas is?

##
a)Oxygen b)Carbon Dioxide c)Ammonia d)Sulphur dioxide

a)Oxygen b)Carbon Dioxide c)Ammonia d)Sulphur dioxide

##### 1 Answer

#### Answer:

I'd go for **(b)**

#### Explanation:

The idea here is that you need to figure out the **molar mass** of the gas by using its density under the given conditions for pressure and temperature.

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

Start by converting the pressure of the gas to *atmospheres* and the temperature to *Kelvin*. You will have

#76 color(red)(cancel(color(black)("cmHg"))) * "1.0 atm"/(76 color(red)(cancel(color(black)("cmHg")))) = "1.0 atm"#

and

#T["K"] = 27^@"C" + 273.15 = "300.15 K"#

Now, you know that the density of the gas, **mass** of the sample, let's say **volume** it occupies,

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

The number of moles of gas can be expressed by using the mass of the sample and the **molar mass** of the gas, let's say

#n = m/M_M#

Plug this into the ideal gas law equation to get

#PV = m/M_M * RT#

Rearrange to isolate the molar mass of the gas

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

This is equivalent to

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

Plug in your values to find

#M_M = "1.80 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")))) * 300.15color(red)(cancel(color(black)("K"))) )/(1.0color(red)(cancel(color(black)("atm"))))#

#M_M = "44 g mol"^(-1) -># rounded totwo sig figs

The closest match is carbon dioxide, which has a molar mass of

#M_ ("M CO"_ 2) = "44.01 g mol"^(-1)#