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

##### 1 Answer

#### Explanation:

You're looking for the mass of potassium sulfite,

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

thenumber of molesof solutethemassof the solvent expressed inkilograms

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 **moles** of solute *per kilogram* of solvent. Since your sample contains

#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**.