# Question 53e9b

Oct 13, 2017

${\text{0.26 mol L}}^{- 1}$

#### Explanation:

The idea here is that you want to express the concentration of this nickel(II) iodide solution in moles per liter, ${\text{mol L}}^{- 1}$.

In order to do that, you need to know two things

• the number of moles of nickel(II) iodide present in your solution, i.e. in $\text{250 mL}$ of this solution
• the number of moles of nickel(II) iodide present in $\text{1 L} = {10}^{3}$ $\text{mL}$ of this solution

To find the number of moles of nickel(II) iodide present in $\text{250 mL}$, you can use the molar mass of the compound.

20.4 color(red)(cancel(color(black)("g"))) * "1 mole NiI"_2/(312.5color(red)(cancel(color(black)("g")))) = "0.06528 moles NiI"_2

You know that $\text{250 mL}$ of this solution contain $0.06528$ moles of nickel(II) iodide, so you can say that $\text{1 L} = {10}^{3}$ $\text{mL}$ of this solution will contain

10^3 color(red)(cancel(color(black)("mL solution"))) * "0.06528 moles NiI"_2/(250color(red)(cancel(color(black)("mL solution")))) = "0.26112 moles NiI"_2#

Since this presents the number of moles of solute present in $\text{1 L}$ of this solution, i.e. the molarity of the solution, you can say that you have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 0.26 mol L}}^{- 1}}}}$

The answer must be rounded to two sig figs, the number of sig figs you have for the volume of the sample.