Question

What is the concentration of NO³⁻ ions in a solution prepared by dissolving 30.0 g of Ca(NO₃)₂ in enough water to produce 300 mL of solution?

a.0.305 M
b.0.609 M
c.0.806 M
d. 1.22 M

Answer 1

The concentration of NO³⁻ ions in the solution is 1.83 M.

Step-by-step explanation:

To find the concentration of NO³⁻ ions in the solution, we first need to calculate the molarity of the Ca(NO₃)₂ solution. We can use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, we convert the mass of Ca(NO₃)₂ to moles:

moles = mass / molar mass

Molar mass of Ca(NO₃)₂ = 164 g/mol

moles = 30.0 g / 164 g/mol = 0.183 moles

Next, we convert the volume of the solution to liters:

volume = 300 mL = 300/1000 L = 0.300 L

Finally, we calculate the molarity:

Molarity = 0.183 moles / 0.300 L = 0.610 M

Since Ca(NO₃)₂ dissociates into three ions: one Ca²⁺ ion and two NO₃⁻ ions, the concentration of NO₃⁻ ions is three times the molarity of the Ca(NO₃)₂ solution:

[NO₃⁻] = 3 * 0.610 M = 1.83 M