# Question 378b3

Apr 25, 2016

$\text{1.27 L}$

#### Explanation:

The idea here is that after all the ice sublimes, the balloon will contain $\text{2.50 g}$ of gaseous carbon dioxide, ${\text{CO}}_{2}$. You can thus use carbon dioxide's molar mass to convert the mass of carbon dioxide to moles.

Once you know how many moles of gaseous carbon dioxide you have in your balloon, you can use the molar volume of a gas at STP to find the volume occupied by your sample.

Now, more often than not, STP (Standard Temperature and Pressure) conditions are defined as a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

This is actually the old definition of STP conditions, but it's still being taught in schools as being accurate.

The thing to remember about STP conditions is that when the pressure of a gas is set at $\text{1 atm}$ and the temperature at ${0}^{\circ} \text{C}$, or $\text{273.15 K}$, one mole of that gas occupies $\text{22.4 L}$.

{(P = "1 atm"), (T = "273.,15 K") :} implies color(blue)(|bar(ul(color(white)(a/a)V = "22.4 mol L"^(-1)color(white)(a/a)|)))

So, use the molar mass of carbon dioxide to determine how many moles you have in that $\text{2.50-g}$ sample

2.50 color(red)(cancel(color(black)("g"))) * overbrace("1 mole CO"_2/(44.01 color(red)(cancel(color(black)("g")))))^(color(Purple)("molar mass of CO"_2)) = "0.0568 moles CO"_2#

So, fi you know that one mole occupies $\text{22.4 L}$ at STP, it follows that this many moles would occupy

$0.0568 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CO"_2))) * overbrace("22.4 L"/(1color(red)(cancel(color(black)("mole CO"_2)))))^(color(purple)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)"1.27 L} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

SIDE NOTE It's worth noting that the current STP conditions are defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

under these conditions, one mole of any ideal gas occupies $\text{22.7 L}$, not $\text{22.4 L}$.