# Question bf70b

Jul 7, 2017

$\text{0.530 g}$

#### Explanation:

The idea here is that you need to use the ideal gas law equation to calculate the number of moles of helium present in the sample, then use the molar mass of helium to convert the number of moles to grams.

The ideal gas law equation looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

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 the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Now, you are told that the partial pressure of helium is equal to $\text{203 mmHg}$.

This value was calculated by taking into account the partial pressures of the other gases present in the mixture and the total pressure of the mixture $\to$ think Dalton's Law of Partial Pressures here.

${P}_{\text{total" = P_ ("CO"_ 2) + P_ "Ar" + P_ ("O"_ 2) + P_ "He}}$

This resulted in

P_ "He" = P_ "total" - (P_ ("CO"_ 2) + P_ "Ar" + P_ ("O"_ 2))

which got you

P_ "He" = "745 mmHg" - ("133 mmHg" + "214 mmHg" + "195 mmHg")

${P}_{\text{He" = "203 mmHg}}$

So, rearrange the ideal gas law equation to solve for $n$

$P V = n R T \implies n = \frac{P V}{R T}$

Plug in your values to find--do not forget to convert the pressure of the gas to atm!

n = (203/760 color(red)(cancel(color(black)("atm"))) * 11.1 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 273 color(red)(cancel(color(black)("K")))) = "0.1323 moles He"#

Since helium has a molar mass of ${\text{4.0026 g mol}}^{- 1}$, you can say that your sample will have a mass of

$0.1323 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles He"))) * "4.0026 g"/(1color(red)(cancel(color(black)("mole He")))) = color(darkgreen)(ul(color(black)("0.530 g}}}}$

The answer is rounded to three sig figs.