Question

Question #53e9b

Answer 1

`"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, `"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 `"250 mL"` of this solution
• the number of moles of nickel(II) iodide present in `"1 L" = 10^3` `"mL"` of this solution

To find the number of moles of nickel(II) iodide present in `"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 `"250 mL"` of this solution contain `0.06528` moles of nickel(II) iodide, so you can say that `"1 L" = 10^3` `"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 `"1 L"` of this solution, i.e. the molarity of the solution, you can say that you have

`color(darkgreen)(ul(color(black)("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.