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
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
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.