Question

Need help calculating molatity if the solution in the lab listed below.

4. Place a weigh boat on the scale. Tare the scale and then weigh out approximately 4.1 g of sodium acetate.

5. Record the exact mass (if you were slightly off from 4.1, that’s fine…just write down the actual value so the math is not incorrect later) of the sodium acetate in your lab notes.

6. Transfer the sodium acetate into a 100 mL beaker.

7. Use a clean and dry 100 mL graduated cylinder to measure and pour approximately 80.0 mL of deionized water into the beaker to dissolve the sodium acetate. Stir the solution until you have a homogenous solution.

8. Pour this solution back into the now empty graduated cylinder, and proceed to Q.S. with dH2O to 100.0 mL. Pour this solution back into its original 100 mL beaker.

9. Calculate the molarity (M, or mol/L) of this sodium acetate solution and label the beaker with the correct concentration and name of the solution. Record the molarity in your lab notes. The molar mass of C₂H₃NaO₂ is 82.03 g/mol.

4.1g Sodium Acetate The molar mass of C₂H₃NaO₂ is 82.03 g/mol.

100ml dH₂O

Correct concentration and name of the solution ____________

Answer 1

The molar mass of sodium acetate (C₂H₃NaO₂) is 82.03 g/mol. This molar mass is 1M in 1 litre flask filled with distilled water. For 100 mL, the molarity or the concentration is 0.1M.

We have taken 4.1g of sodium acetate and made up the solution in a 100 mL beaker. Therefore, the concentration of sodium acetate is :

4.1g$\div$82.03g/mol$\times$0.1M = 0.005M

The name of the solution is 0.005M sodium acetate solution.