What is the percent by mass of KBr if 58.2 grams are dissolved in 344 grams of water?

The idea here is that you need to determine how many grams of potassium bromide will be present for every #"100. g"# of this solution, which is what the solution's percent concentration by mass represents.

If you dissolve #"58.2 g"# of potassium bromide in #"344 g"# of water, the total mass of the solution will be

#"58.2 g + 344 g = 402.2 g"#

So this solution contains #"58.2 g"# of potassium bromide in #"402.2 g"# of the solution, which means that in #"100. g"# of this solution you will have

#100. color(red)(cancel(color(black)("g solution"))) * "58.2 g KBr"/(402.2 color(red)(cancel(color(black)("g solution")))) = "14.5 g KBr"#

You can thus say that the solution's percent concentration by mass is equal to

#color(darkgreen)(ul(color(black)("% m/m = 14.5% KBr")))#

The answer is rounded to three sig figs, the number of sig figs you have for your figures.

#"%(m/m)"=("mass of solute")/("mass of solution")xx100#

The mass of the solution is the mass of the solute and solvent.

The given mass of the solute #"KBr"# is #"58.2 g"#.

The mass of the solution is #"58.2 g KBr + 344 g H"_2"O"="404.2 g"#

Plug in the known values and solve.

#"%(m/m)"=(58.2"g KBr")/(404.2"g solution")xx100=14.4%#