Question #53e9b

1 Answer
Oct 13, 2017

Answer:

#"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.