# Question 428df

Jun 21, 2017

$V = 56.0$ $\text{L}$

#### Explanation:

To solve this problem, we can first convert the given mass of $\text{He}$ to moles, using its molar mass ($4.00 \text{g"/"mol}$, as seen from a periodic table):

10.0cancel("g He")((1color(white)(l)"mol He")/(4.00cancel("g He"))) = 2.50 $\text{mol He}$

At standard temperature and pressure conditions ($1$ bar and ${0}^{\text{o""C}}$), one mole of an ideal gas has a volume of $22.41$ $\text{L}$.

Therefore, using dimensional analysis again,

2.50cancel("mol He")((22.41color(white)(l)"L")/(1cancel("mol He"))) = color(red)(56.0 color(red)("L"

Thus, $10.0$ grams of helium occupies a volume of roughly color(red)(56.0 sfcolor(red)("liters"#.