Question #678e3

The thing to remember about a solution's molarity is that it tells you how many moles of solute are present for every #"1 L"# of solution.

In this case, a #"1.5-M"# potassium bromide solution will contain #1.5# moles of potassium bromide, the solute, for every #"1 L"# of solution.

#color(blue)(ul(color(black)("1.5 M KBr = 1.5 moles KBr in every 1 L of KBr solution")))#

Now, you know that

#1 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "1000 mL"#

This means that your goal here is to figure out how many moles of potassium bromide you get in #1/10"th"# of a liter of #"1.5 M"# potassium bromide solution, i.e. in #"100 mL"# of solution.

To do that, you can use the molarity of the solution as a conversion factor

#100 color(red)(cancel(color(black)("mL solution"))) * overbrace("1.5 moles KBr"/(1000color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 1.5 M KBr solution")) = color(darkgreen)(ul(color(black)("0.15 moles KBr")))#

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the volume of the sample.