# Question ccb82

Mar 28, 2017

Here's what I got.

#### Explanation:

The density of the solution can be found by dividing its total mass by its total volume.

You can safely assume that dissolving $\text{2 g}$ of sodium chloride in $\text{500 mL}$ of water will not change the volume of the solution, so right from the start, you know that

${V}_{\text{solution" = V_"water" = "500 mL}}$

Now, you know the mass of sodium chloride, which is the solute, so you must figure out the mass of the water, which is the solvent.

Using the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{1 mL" = "1 cm}}^{3}}}}$

and that water has a density of ${\text{1 g cm}}^{- 3}$, you can say that $\text{500 mL}$ of water have a mass of

500 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("mL")))) * "1 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "500 g"#

Therefore, the total mass of the solution will be

${m}_{\text{total" = m_"NaCl" + m_"water}}$

${m}_{\text{total" = "2 g + 500 g}}$

${m}_{\text{total" = "502 g}}$

The density of the solution, ${\rho}_{\text{sol}}$, will be equal to

${\rho}_{\text{sol" = m_"total"/V_"solution}}$

Plug in your values to find

${\rho}_{\text{sol" = "502 g"/"500 mL" = "1.004 g mL}}^{- 1}$

Now, I'll leave the answer rounded to four sig figs, but keep in mind that it should be rounded to one significant figure.

So the correct answer here would be

${\rho}_{\text{sol" = "1 g mL}}^{- 1}$