# Question c2a76

Jan 7, 2017

Here's what I got.

#### Explanation:

The first thing you need to do here is to figure out the number of moles of barium nitrate, "Ba"("NO"_3)_2, that can be dissolved in $\text{1 L}$ of solution at ${25}^{\circ} \text{C}$.

To do that, use the molar mass of the compound

130.6 color(red)(cancel(color(black)("g"))) * ("1 mole Ba"("NO"_3)_2)/(261.34color(red)(cancel(color(black)("g")))) = "0.49973 moles Ba"("NO"_3)_2

Now, you can use this value to write the solubility of barium nitrate in moler per liter, ${\text{mol L}}^{- 1}$.

${\text{130.6 g L"^(-) = "0.49973 mol L}}^{- 1}$

Now, $1$ mole of barium nitrate contains

• one mole of barium cations, $1 \times {\text{Ba}}^{2 +}$
• two moles of nitrate anions, $2 \times {\text{NO}}_{3}^{-}$

This means that if you dissolve $0.49973$ moles of barium nitrate to make $\text{1 L}$ of solution at ${25}^{\circ} \text{C}$, you will end up with

0.49973 color(red)(cancel(color(black)("moles Ba"("NO"_3)_2))) * "1 mole Ba"^(2+)/(1color(red)(cancel(color(black)("mole Ba"("NO"_3)_2)))) = "0.50 moles Ba"^(2+)

0.49973 color(red)(cancel(color(black)("moles Ba"("NO"_3)_2))) * "2 mole NO"_3^(-)/(1color(red)(cancel(color(black)("mole Ba"("NO"_3)_2)))) = "1.0 moles NO"_3^(-)

The total number of moles of ions present in solution will be

"0.50 moles Ba"^(2+) + "1.0 moles NO"_3^(-) = color(darkgreen)(ul(color(black)("1.5 moles ions")))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the volume of the solution.