# The density of #"CO"_2# at #0^@ "C"# is #"1.98 g/L"#. How many mols are found in #"88.0 L"# of #"CO"_2# at this temperature and #"1 atm"#?

##### 2 Answers

#### Answer:

#### Explanation:

The thing to remember about **Standard Temperature and Pressure**, **STP**, is that when an ideal gas is being kept under these conditions for *pressure* and *temperature*, **one mole** of this gas occupies

The volume occupied by one mole of an ideal gas at *any* conditions for pressure and temperature is called the **molar volume**.

Under STP conditions, which are **currently** defined as a pressure of **molar volume of a gas at STP**, which is equal to

So, you can use the molar volume of a gas at STP as a *conversion factor* to figure out how many moles of carbon dioxide,

#88.0 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas aat STP")) = "3.87665 moles"#

Rounded to three **sig figs**, the answer will be

#"volume of CO"_2 = color(green)(|bar(ul(color(white)(a/a)"3.88 moles"color(white)(a/a)|)))#

**SIDE NOTE** *More often than not, you'll see STP conditions being defined as a pressure of* *and a temperature of*

*Under these conditions for pressure and temperature, one mole of any ideal gas occupies*

*If this is the value given to you for the molar volume of a gas at STP, simply redo the calculation using* *instead of*

Here's an alternative way to do this, without requiring the assumption that

If you use the density of

The main catch with using the ideal gas law is that it assumes all gases have the *same* **molar volume** at STP, and of course, the "more realistic" gases don't.

When you use the experimental density and *convert* that to the **molar density**, it circumvents that assumption because you'd be using all experimental data. Here's how you do it:

#1/(barV) = barrho = n/V = m/(VM_r) = \mathbf((rho)/M_r)# where:

#V# is volume in#"L"# #n# is#"mol"# s#M_r# is relative molar mass in#"g/mol"# #m# is mass in#"g"# #rho# is the mass density in#"g/mol"#

The molar density has units of

#"mol CO"_2 = ("1.98" cancel"g")/(cancel"L")xx("mol")/(44.009 cancel"g") xx 88.0 cancel"L"#

# = color(blue)("3.92 mol CO"_2)#

And this is larger but somewhat close to Stefan's answer of

This answer agrees with that statement because there are a greater number of