# Question 50761

Dec 14, 2017

$30.072 L$

#### Explanation:

We know that,

The atomic mass of $H e$ is $4 g = 1 m o l$

At $S T P ,$

the molar volume of every gasses of $1 m o l$ is $22.4 L$

Then,at $S T P$

The volume of $4 g H e$ is$22.4 L$

$\therefore$The volume of $5.37 g H e$ is
$= \frac{5.37 g \times 22 , 4 L}{4 g} = 30.072 L$

Dec 14, 2017

See the simple steps to take in solving the question above;

#### Explanation:

We are dealing on two given information..

Volume and Mass..

Using this we can get the required Volume..

Recall;

$n = \frac{v}{m V} = \frac{m}{m M}$

Where;

$n = \text{no of moles}$

$v = \text{volume}$

$m = \text{mass}$

$m V = \text{molar volume}$

$m M = \text{molar mass}$

So we have;

$\frac{v}{m V} = \frac{m}{m M}$

v = ?#

$m V = 22.4 {\mathrm{dm}}^{3} m o {l}^{-} 1 \mathmr{and} L m o {l}^{-} 1$

N/B $\Rightarrow {\mathrm{dm}}^{3} = L$

$m = 5.37 g$

$m M = 4 g m o {l}^{-} 1$

Substitute the values into the fomula

$\frac{v}{22.4 {\mathrm{dm}}^{3} m o {l}^{-} 1} = \frac{5.37 g}{4 g m o {l}^{-} 1}$

$\frac{v}{22.4 {\mathrm{dm}}^{3} \cancel{m o {l}^{-} 1}} = \frac{5.37 \cancel{g}}{4 \cancel{g} \cancel{m o {l}^{-} 1}}$

$\frac{v}{22.4 {\mathrm{dm}}^{3}} = \frac{5.37}{4}$

Cross multiplying..

$v \times 4 = 22.4 {\mathrm{dm}}^{3} \times 5.37$

$4 v = 22.4 {\mathrm{dm}}^{3} \times 5.37$

$4 v = 120.288 {\mathrm{dm}}^{3}$

Divide both sides by $4$

$\frac{\cancel{4} v}{\cancel{4}} = \frac{120.288 {\mathrm{dm}}^{3}}{4}$

$v = \frac{120.288 {\mathrm{dm}}^{3}}{4}$

$v = 30.072 {\mathrm{dm}}^{3} \mathmr{and} 30.072 L$