Question

A 5.00 mL sample of a solution of KCl (Mwt=74.6 g/mol) in water (Mwt=18.01 g/mol) weighs 5.20 g. The water is evaporated off leaving 0.26 g of residue. What is the mass of solution that would contain 1.00 g of KCl?

Answer 1

`"20. g"`

Explanation:

The first thing you need to do here is to figure out the density of the solution, `rho`, by using its mass, `m`, and its volume, `V`

`color(blue)(ul(color(black)(rho = m/V)))`

In your case, you will have

`rho = "5.20 g"/"5.00 mL" = "1.04 g mL"^(-1)`

Now, the key thing to keep in mind here is that solutions are homogeneous mixtures, which implies that they have the same composition throughout.

This allows you to use the known composition of the initial solution, which contains `"0.26 g"` of potassium chloride in `"5.00 mL"` of solution, as a conversion factor in order to determine the volume of solution that will contain `"1.00 g"` of potassium chloride.

`1.00 color(red)(cancel(color(black)("g KCl"))) * overbrace("5.00 mL solution"/(0.26color(red)(cancel(color(black)("g KCl")))))^(color(blue)("the known composition of the solution")) = "19.23 mL solution"`

Finally, use the density of the solution to convert the volume to mass

`19.23 color(red)(cancel(color(black)("mL solution"))) * overbrace("1.04 g"/(1color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 1.04 g mL"^(-1))) = color(darkgreen)(ul(color(black)("20. g")))`

The answer is rounded to two sig figs, the number of sig figs you have for the mass of potassium chloride.