The solubility of #Ba(NO_3)_2# is 130.5 g/L at #0^o C#. How many moles of dissolved salt are present in 4.0 L of a saturated solution of #Ba(NO_3)_2# at #0^oC#?

The problem provides you with the solubility of barium nitrate, #"Ba"("NO"_3)_2#, in grams per liter, #"g L"^(-1)#, in water at #0^@"C"#, and asks you to figure out how many moles of barium nitrate can be dissolved in #"4.0 L"# of aqueous solution at #0^@"C"#.

The first thing to do here is recognize the fact that the solubility of barium nitrate tells you how many grams of the salt can be dissolved in one liter of water at #0^@"C"# in order to make a saturated solution.

In this case, barium nitrate is said to have a solubility of #"130.5 g L"^(-1)#, which tells you that at #0^@"C#", you can only hope to dissolve #"130.5 g"# of barium nitrate per liter of water.

Use the compound's molar mass to convert the mass of barium nitrate to moles

#130.5 color(red)(cancel(color(black)("g"))) * ("1 mole Ba"("NO"_3)_2)/(261.34color(red)(cancel(color(black)("g")))) = "0.4993 moles Ba"("NO"_3)_2#

Now, you know that #"1 L"# of water can dissolve #0.4993# moles of barium nitrate at #0^@"C"#, which means that #"4.0 L"# of water can dissolve

#4.0 color(red)(cancel(color(black)("L water"))) * ("0.4993 moles Ba"("NO"_3)_2)/(1color(red)(cancel(color(black)("L water")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.0 moles Ba"("NO"_3)_2)color(white)(a/a)|)))#

The answer is rounded to two sig figs, the number of sig figs you have for the volume of water.