Question

Question #e164a

Answer 1

`"0.005 moles"`

Explanation:

The thing to remember about a solution's molarity is that you can use it as a conversion factor to go from moles of solute to volume of solution or vice versa.

As you know, a solution's molarity tells you the number of moles of solute present for every `"1 dm"^3` of solution.

In your case, a `"0.1 M"` solution will contain `0.1` moles of solute for every

`1 color(red)(cancel(color(black)("dm"^3))) * (10^3"cm"^3)/(1color(red)(cancel(color(black)("dm"^3)))) = 10^3"cm"^3`

of solution. This means that `"50 cm"^3` of this solution will contain

`50 color(red)(cancel(color(black)("cm"^3color(white)(.)"solution"))) * overbrace("01 moles solute"/(10^3color(red)(cancel(color(black)("cm"^3color(white)(.)"solution")))))^(color(blue)("= 0.1 M")) = color(darkgreen)(ul(color(black)("0.005 moles solute")))`

The answer is rounded to one significant figure.