# Question df940

Feb 21, 2017

#### Answer:

$\text{38.6 g}$

#### Explanation:

The tricky part here is actually to get the conditions for Room Temperature and Pressure (RTP) right. Since you didn't provide more info about the RTP conditions given to you, I'll work with a temperature of ${25}^{\circ} \text{C}$ and a pressure of $\text{1 atm}$.

Now, the idea is that you can use the ideal gas law equation to find the molar volume of a gas at RTP, i.e. the volume occupied by $1$ mole of gas under these conditions.

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

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

Rearrange the ideal gas law equation

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

Plug in your values -- do not forget to convert the temperature to Kelvin!

V/n = (0.0821(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("atm"))))

$\frac{V}{n} = {\text{24.45 L mol}}^{- 1}$

This tells you that under RTP conditions, $1$ mole of any ideal gas occupies $\text{24.45 L}$. Since you know that your sample occupies

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 dm"^3 = "1 L}}}}$

you can say that it contains

1 color(red)(cancel(color(black)("L"))) * "1 mole gas"/(24.45color(red)(cancel(color(black)("L")))) = "0.04090 moles gas"#

Finally, you know that this sample has a mass of $\text{1.58 g}$, which means that the mass of $1$ mole will be equal to

$1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mole"))) * "1.58 g"/(0.04090color(red)(cancel(color(black)("moles")))) = color(darkgreen)(ul(color(black)("38.6 g}}}}$

I'll leave the answer rounded to three sig figs.

Notice that you've essentially determined the molar mass of the gas, i.e. the mass of $1$ mole of this gas.

SIDE NOTE Notice that you can get to the result a bit faster by rearranging the ideal gas law equation as

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

This will give you the number of moles present in one liter of gas kept under RTP conditions. Once again, you will have

$\frac{n}{V} = \left(1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K}}}}\right)$

$\frac{n}{V} = {\text{0.04090 mol L}}^{- 1}$

Since you know that your sample occupies $\text{1 L}$, you can say that it contains $0.04090$ moles of gas.