# Question e164a

Mar 13, 2017

$\text{0.005 moles}$

#### 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 ${\text{1 dm}}^{3}$ of solution.

In your case, a $\text{0.1 M}$ 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 ${\text{50 cm}}^{3}$ of this solution will contain

$50 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{cm"^3color(white)(.)"solution"))) * overbrace("01 moles solute"/(10^3color(red)(cancel(color(black)("cm"^3color(white)(.)"solution")))))^(color(blue)("= 0.1 M")) = color(darkgreen)(ul(color(black)("0.005 moles solute}}}}$

The answer is rounded to one significant figure.