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Sagot :
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:
For solution 1:
Volume = 45.3 mL = 45.3 / 1000 = 0.0453 L
Molarity = 0.549 M
Mole of Ca(NO₃)₂ =?
Mole = Molarity x Volume
Mole of Ca(NO₃)₂ = 0.549 × 0.0453
Mole of Ca(NO₃)₂ = 0.0249 mole
For solution 2:
Volume = 75.05 mL = 75.05 / 1000 = 0.07505 L
Molarity = 1.321 M
Mole of Ca(NO₃)₂ =?
Mole = Molarity x Volume
Mole of Ca(NO₃)₂ = 1.321 × 0.07505
Mole of Ca(NO₃)₂ = 0.0991 mole
- 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
Total mole = 0.124 mole
- 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
Total Volume = 0.12035 L
- 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
Concentration =?
Concentration = mole / Volume
Concentration = 0.124 / 0.12035
Concentration = 1.03 M
Therefore, the concentration of the final solution of Ca(NO₃)₂ is 1.03 M
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