# If the mole fraction of #"K"_2"SO"_3# in a solution is 0.0328, what is the percent by mass of #"K"_2"SO"_3#?

##### 1 Answer

#### Answer:

#### Explanation:

The **mole fraction** of potassium sulfite,

By definition, this mole fraction is equal to the number of moles of potassium sulfite divided by the **total number of moles** present in solution.

A useful thing to keep in mind here is that because your solution has two components, i.e. potassium sulfite as the *solute*and water as the *solvent*, their respective **mole fractions** must add up to give

This means that the mole fraction of water will be

#1 - 0.0328 = 0.9672#

Now, to make the calculations easier, let's take a *number of moles* of potassium sulfite and

You can say that

#(x color(red)(cancel(color(black)("moles"))))/((x+y)color(red)(cancel(color(black)("moles")))) = 0.0328 -># themole fractionof#"K"_2"SO"_3#

Use the **molar masses** of potassium sulfite and water to write

#x color(red)(cancel(color(black)("moles K"_2"SO"_3))) * "158.26 g"/(1color(red)(cancel(color(black)("mole K"_2"SO"_3)))) = (158.26 * x)" g K"_2"SO"_3#

#y color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = (18.015 * y)" g H"_2"O"#

These two masses must add up to give the **total mass** of the sample

#(158.26 * x) color(red)(cancel(color(black)("g"))) + (18.015 * y)color(red)(cancel(color(black)("g"))) = 100color(red)(cancel(color(black)("g")))#

#158.26 * x + 18.015 * y = 100#

Use this equation to find

#x = (100 - 18.015 * y)/158.26#

Plug this into the expression you have for the *mole fraction* of potassium sulfite to get

#x = 0.0328 * (x+y)#

#0.9672 * x = 0.0328 * y#

#0.9672 * [(100 - 18.015 * y)/158.26] = 0.0328 * y#

This will be equivalent to

#96.72 - 17.424 * y = 5.191 * y#

#22.615 * y = 96.72 implies y = 96.72/22.615 = 4.277#

The value of

#x = (100 - 18.015 * 4.277)/158.26 = 0.145#

You can say that the

#0.145 color(red)(cancel(color(black)("moles K"_2"SO"_3))) * "158.26 g"/(1color(red)(cancel(color(black)("mole K"_2"SO"_3)))) = "22.95 g K"_2"SO"_3#

Therefore, the **percent concentration by mass** of potassium sulfite will be

#"% K"_2"SO"_3 = (22.95 color(red)(cancel(color(black)("g"))))/(100color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(23.0%)color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.