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
Aug 25, 2016

Answer:

#"23.0% K"_2"SO"_3#

Explanation:

The mole fraction of potassium sulfite, #"K"_2"SO"_3#, in the solution is said to be equal to #0.0328#.

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 soluteand water as the solvent, their respective mole fractions must add up to give #1#.

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 #"100-g"# sample of this solution. Let's say that #x# represents the number of moles of potassium sulfite and #y# represents the number of moles of water.

You can say that

#(x color(red)(cancel(color(black)("moles"))))/((x+y)color(red)(cancel(color(black)("moles")))) = 0.0328 -># the mole fraction of #"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# will thus be

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

You can say that the #"100-g"# sample contains

#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.