Question

a student combined 45.3 ml of a 0.549 m calcium nitrate ca(no₃)₂ solution with 75.05 ml of a 1.321 m ca(no₃)₂ solution. calculate the concentration of the final solution.

Answer 1

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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