The concentration of the final solution of Ca(NO₃)₂ made by combining 45.3 mL of a 0.549 M Ca(NO₃)₂ solution with 75.05 mL of a 1.321 M Ca(NO₃)₂ solution is 1.03 M
We'll begin by calculating the number of mole of Ca(NO₃)₂ in each solution. This can be obtained as follow:
<h3>For solution 1:</h3>
Volume = 45.3 mL = 45.3 / 1000 = 0.0453 L
Molarity = 0.549 M
<h3>Mole of Ca(NO₃)₂ =?</h3>
Mole = Molarity x Volume
Mole of Ca(NO₃)₂ = 0.549 × 0.0453
<h3>Mole of Ca(NO₃)₂ = 0.0249 mole</h3>
<h3>For solution 2:</h3>
Volume = 75.05 mL = 75.05 / 1000 = 0.07505 L
Molarity = 1.321 M
<h3>Mole of Ca(NO₃)₂ =?</h3>
Mole = Molarity x Volume
Mole of Ca(NO₃)₂ = 1.321 × 0.07505
<h3>Mole of Ca(NO₃)₂ = 0.0991 mole</h3>
- Next, we shall determine the total mole of Ca(NO₃)₂ in the final solution. This can be obtained as follow:
Mole of Ca(NO₃)₂ in solution 1 = 0.0249 mole
Mole of Ca(NO₃)₂ in solution 2 = 0.0991 mole
Total mole = 0.0249 + 0.0991
<h3>Total mole = 0.124
mole</h3>
- Next, we shall determine the total volume of the final solution.
Volume of solution 1 = 0.0453 L
Volume of solution 2 = 0.07505 L
Total Volume = 0.0453 + 0.07505
<h3>Total Volume = 0.12035 L</h3>
- Finally, we shall determine the concentration of the final solution. This can be obtained as follow:
Total Volume = 0.12035 L
Total mole = 0.124 mole
<h3>Concentration =?</h3>
Concentration = mole / Volume
Concentration = 0.124 / 0.12035
<h3>Concentration = 1.03 M</h3>
Therefore, the concentration of the final solution of Ca(NO₃)₂ is 1.03 M
Learn more: brainly.com/question/25469095
