# 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?

##### 1 Answer

#### Answer:

#### Explanation:

The first thing you need to do here is to figure out the **density** of the solution,

#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 **conversion factor** in order to determine the *volume* of solution that will contain

#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.