# Question 1cb83

Feb 15, 2018

Here's what I got.

#### Explanation:

The trick here is to realize that under STP conditions, which are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$, $1$ mole of any ideal gas occupies $\text{22.723 L}$--this is known as the molar volume of agas at STP.

Now, you know that at STP, this gas has a density of ${\text{1.7824 g L}}^{- 1}$. This tells you that at a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$, $\text{1.7824 g}$ of this gas occupies exactly $\text{1 L}$.

Use the molar volume of a gas at STP to find the number of moles present in the sample

1 color(red)(cancel(color(black)("L"))) * "1 mole gas"/(22.723color(red)(cancel(color(black)("L")))) = "0.0440083 moles gas"

To find the molar mass of the gas, you need to find the mass of exaftly $1$ mole. Since you know that $0.0440083$ moles have a mass of $\text{1.7824 g}$, you can say that

1 color(red)(cancel(color(black)("mole"))) * "1.7824 g"/(0.0440083color(red)(cancel(color(black)("moles")))) = "40.501 g"#

Therefore, you can say that the molar mass of the gas is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molar mass = 40.501 g mol}}^{- 1}}}}$

The answer is rounded to five sig figs, the number of sig figs you have for the density of the gas at STP.

$\textcolor{w h i t e}{a}$
SIDE NOTE More often than not, the molar volume of a gas at STP is given as ${\text{22.414 mol L}}^{- 1}$, the value that corresponds to a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

If that's the value given to you, make sure to redo the calculations using ${\text{22.414 L mol}}^{- 1}$ instead of ${\text{22.723 mol L}}^{- 1}$.