# How many moles of argon atoms are present in 11.2 L of argon gas at STP?

Jan 8, 2017

What is the molar volume at "STP"?

#### Explanation:

Well, the molar volume at $\text{STP}$ is $22.4 \cdot L$. You have to be careful to quote this now, because different curricula have different standards, and some quote standard pressure as $100 \cdot k P a$ not $\text{1 atmosphere}$.

So you have a volume of $11.2 \cdot L$; given the molar volume, and this represents, (11.2*cancelL)/(22.4*cancelL*mol^-1)=??mol.

We note that $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$, as required for a molar quantity.

Jan 8, 2017

Using the current recommended STP values of $\text{0"^@"C}$ and $\text{100 kPa}$, $\text{11.2 L}$ of Ar atoms contain $\text{0.493 mol Ar}$.

Using the older standard, as indicated in the explanation, $\text{11.2 L}$ of Ar atoms contain $\text{0.500 mol}$.

#### Explanation:

Current values for STP are ${0}^{\circ} \text{C}$ or $\text{273.15 K}$ for temperature, and pressure of $\text{10"^5color(white)(.) "Pa}$, which is usually written as $\text{100 kPa}$, or $\text{1 bar}$.

At these current values, the molar volume of a gas is $\text{22.710 L/mol}$.

In order to determine the number of moles of the given volume of Ar, divide the given volume by the molar volume of an ideal gas, $\text{22.710 L/mol}$.

$\left(11.2 \text{L")/(22.710"L"/"mol}\right)$

Simplify by multiplying by the inverse of the molar volume.

$11.2 \cancel{\text{L"xx(1"mol")/(22.710cancel"L")="0.493 mol Ar}}$

Using the older STP values of $\text{0"^@"C}$ or $\text{273.15 K}$, and pressure of $\text{1 atm}$, molar volume is $\text{22.414 L/mol}$.

$11.2 \cancel{\text{L"xx(1"mol")/(22.414cancel"L")="0.500 mol Ar}}$