<u>Answer:</u> The molarity of
ions in the solution is 0.299 M
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
- <u>For chromium (III) acetate:</u>
Molarity of chromium (III) acetate solution = 0.234 M
Volume of solution = 26.2 mL
Putting values in equation 1, we get:

1 mole of chromium (III) acetate
produces 1 mole of chromium
ions and 3 moles of acetate
ions
Moles of 
- <u>For chromium (III) nitrate:</u>
Molarity of chromium (III) nitrate solution = 0.461 M
Volume of solution = 10.7 mL
Putting values in equation 1, we get:

1 mole of chromium (III) nitrate
produces 1 mole of chromium
ions and 3 moles of nitrate
ions
Moles of 
- <u>For chromium cation:</u>
Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles
Total volume of solution = [26.2 + 10.7] = 36.9 mL
Putting values in equation 1, we get:
ions in the solution is 0.299 M