# Question #af574

##### 1 Answer

#### Explanation:

The idea here is that you need to use the known composition of the solution to figure out the mass of glucose, the solute, present in exactly **percent concentration by mass**.

To do that, you can use the fact that solutions are **homogeneous mixtures**, which implies that they have the same composition throughout.

So if you dissolve **solution** will be equal to

#overbrace("7.79 g")^(color(blue)("mass of solute")) + overbrace("237.7 g")^(color(blue)("mass of solvent")) = overbrace("245.5 g")^(color(blue)("mass of solution"))#

So, you know that this solution contains

#100 color(red)(cancel(color(black)("g solution"))) * "7.79 g glucose"/(245.5color(red)(cancel(color(black)("g solution")))) = "3.17 g glucose"#

This means that the solution's **percent concentration by mass** will be equal to

#color(darkgreen)(ul(color(black)("% m/m = 3.17% glucose")))#

The answer is rounded to three **sig figs**, the number of sig figs you have for the mass of glucose.