Question #c2a76

1 Answer
Jan 7, 2017

Answer:

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.