# Question 78d59

Nov 14, 2016

${\text{0.002271 moles K"_2"Cr"_2"O}}_{7}$

#### Explanation:

By defition, a $\text{0.1885-M}$ solution of potassium dichromate, ${\text{K"_2"Cr"_2"O}}_{7}$, contains $0.1885$ moles of solute for every liter of solution.

Since you know that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 L" = 10^3"mL}}}}$

you can say that your solution will contain

"0.1885 M K"_2"Cr"_2"O"_7 = ("0.1885 moles K"_2"Cr"_2"O"_7)/(10^3"mL solution")#

You can thus use the molarity of the solution as a conversion factor to take you from volume of solution to moles or vice versa.

$12.05 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{mL"))) * ("0.1885 moles K"_2"Cr"_2"O"_7)/(10^3color(red)(cancel(color(black)("mL")))) = color(darkgreen)(ul(color(black)("0.002271 moles K"_2"Cr"_2"O}}_{7}}}}$