Answer:
[Ag+] = [NO3-] = 0.700M
0.555M = [Na+] = [I-]
Explanation:
To solve this question we need to find the moles of sodium iodide, NaI, using its molar mass -. With the moles and the volume we can find the molarity of Na+ and I-. The molarity of the ions of silver nitrate, AgNO3 doesn't change because we are assuming the volume doesn't change:
Molarity Ag⁺ = Molarity NO₃⁻ = 0.700M
Moles NaI -Molar mass: 149.89g/mol-
20.8g NaI * (1mol/149.89g) = 0.0139 moles NaI
Molarity:
0.0139 moles NaI / 0.250L = 0.555M = [Na+] = [I-]
