# An empty vial weighs 50.90 g. If the vial weighs 439.31 g when filled with liquid mercury (d = 13.53 g/cm^3), what volume of mercury is in the vial?

Sep 10, 2016

${\text{28.71 cm}}^{3}$

#### Explanation:

Your strategy here will be to use the mass of the empty vial and the mass of the vial when filled with mercury to find the mass of mercury added to the vial.

Once you know the mass of mercury, you can use its density to find the volume of mercury present in the vial, i.e. the volume of the vial.

So, you know that the vial + mercury has a mass of

$\text{vial " + " mercury" = "439.31 g}$

The mass of the vial is

$\text{vial " = " 50.90 g}$

This means that the mass of the mercury will be

$\text{mercury" = "439.31 g" - "50.90 g" = "388.41 g}$

Now, you know that mercury has a density of ${\text{13.53 g/cm}}^{3}$. What this means is that ${\text{1 cm}}^{3}$ of mercury has a mass of $\text{13.53 g}$.

In your case, the volume of the sample will be

388.41 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(13.53color(red)(cancel(color(black)("g")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("28.71 cm"^3)color(white)(a/a)|)))

The answer is rounded to four sig figs, the number of sig figs you have for the mass of the empty vial.