Question #e164a

1 Answer
Mar 13, 2017

#"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.