How many grams of potassium sulfite are required to dissolve in 872g of water to make a 0.128m solution?

1 Answer
Jul 1, 2016

#"17.7 g"#


You're looking for the mass of potassium sulfite, #"K"_2"SO"_3"#, needed to make a #"0.128 m"#, or #"0.128 molal"#, solution.

Now, molality is used to express the concentration of a solution in terms of how many moles of solute it contains per kilogram of solvent.

This means that in order to find a solution's molality, you need to know

  • the number of moles of solute
  • the mass of the solvent expressed in kilograms

In your case, you already know the mass of the solvent in grams, so the very first thing to do here is convert it to kilograms

#872 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.872 kg"#

Now, a #"0.128 m"# solution contains #0.128# moles of solute per kilogram of solvent. Since your sample contains #"0.872 kg"# of solvent, it follows that it will contain

#0.872 color(red)(cancel(color(black)("kg solvent"))) * overbrace(("0.128 moles K"_2"SO"_3)/(1color(red)(cancel(color(black)("kg solvent")))))^(color(blue)("= 0.128 m")) = "0.1116 moles K"_2"SO"_3#

All you have to do now is use the molar mass of potassium sulfite to figure out how many grams would contain that many moles

#0.1116 color(red)(cancel(color(black)("moles K"_2"SO"_3))) * "158.26 g"/(1color(red)(cancel(color(black)("mole K"_2"SO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("17.7 g")color(white)(a/a)|)))#

The answer is rounded to three sig figs.