Question #af574

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 #"100 g"# of the solution #-># this is the solution's 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 #"7.79 g"# of glucose in #"237.7 g"# of water, you can say that the mass of the 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 #"7.79 g"# of solute in #"245.5 g"# of solution, so you can say that #"100 g"# of this solution will contain

#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.