Question #f11c8

The trick here is to realize that the mass of added sugar will increase the total mass of sugar and the total mass of the solution.

Your initial solution is #10%# sugar by mass, which means that every #"100 g"# of this solution contain #"10 g"# of sugar. You can thus say that the initial solution contains

#75 color(red)(cancel(color(black)("g solution"))) * "10 g sugar"/(100color(red)(cancel(color(black)("g solution")))) = "7.5 g sugar"#

Let's say that you are going to add #x# #"g"# of sugar to this solution to get its percent concentration to #15%#. At this point, the solution will contain #"15 g"# of sugar for every #"100 g"# of solution.

The mass of the solution will be

#"75 g" + x color(white)(.)"g" = (75 + x)# #"g"#

This means that the second solution will contain

#(75 + x) color(red)(cancel(color(black)("g solution"))) * "15 g sugar"/(100color(red)(cancel(color(black)("g solution")))) = 0.15 * (75 +x)# #"g sugar"#

But you already know that the mass of sugar will also be equal to

#"7.5 g" + xcolor(white)(.)"g" = (7.5 + x)# #"g"#

You can thus say that

#0.15 * (75 + x) color(red)(cancel(color(black)("g"))) = (7.5 + x)color(red)(cancel(color(black)("g")))#

Solve this for #x# to get

#11.25 + 0.15x = 7.5 + x#

#3.25 = 0.85x implies x = 3.25/0.85 = 4.41#

Therefore, you can say that if you add #"4.41 g"# of sugar to #"75 g"# of #10%# by mass sugar solution, you will get #"79.41 g"# of #15%# by mass sugar solution.