In this problem, you need to find two things: the moles of solute (NO₃⁻) and the volume of the total solution.

First, find the moles of NO₃⁻ we have in each solution.

n₁ = (0.263 L)(0.167 M KNO₃) = 0.043921 mol KNO₃

Now, since there is one mole of NO₃⁻ per one mole of KNO₃, we can say easily say that the number of moles of NO₃⁻ is also 0.043921 mol.

n₂ = (0.311 L)(0.404 M Mg(NO₃)₂) = 0.125644 mol Mg(NO₃)₂

IMPORTANT: since there are TWO moles per one mole of Mg(NO₃)₂, we have to say that the number of moles of NO₃⁻ is DOUBLED the number of moles in Mg(NO₃)₂. Thus, it is 0.251288 mol NO₃⁻

Now we need to find the TOTAL number of moles of NO₃⁻.

n = 0.043921 mol + 0.251288 mol = 0.295209 mol NO₃⁻

Next, find the total volume. (pretty easy...)

V = 0.263 L + 0.311 L + 0.713 L = 1.287 L

Finally, use all the info we calculated to find the concentration of NO₃⁻ in the resulting solution.

M = (0.295209 mol) / (1.287 L) = 0.229 M NO₃⁻

Note: Notice how the final answer has only 3 sig figs. You do not round until the end. Remember that :)

Thus, the NO₃⁻ concentration of the resulting solution is 0.229 M NO₃⁻