Question

Question #c2a76

Answer 1

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 `"1 L"` of solution at `25^@"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, `"mol L"^(-1)`.

`"130.6 g L"^(-) = "0.49973 mol L"^(-1)`

Now, `1` mole of barium nitrate contains

• one mole of barium cations, `1 xx "Ba"^(2+)`
• two moles of nitrate anions, `2 xx "NO"_3^(-)`

This means that if you dissolve `0.49973` moles of barium nitrate to make `"1 L"` of solution at `25^@"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.