TUTORIAL – Lab: Experiment 8 Practice Questions

For Experiment 8, you will need to complete some calculations for Part B and C before going to the lab, otherwise, you will not have time to finish the lab on time.

Answer to Question 1:

The question tells us the following information:

1. The concentration of the solution we want to make. It is 0.0380 M CuSO₄.
2. The volume of the solution we want to make. It is 250.0 mL (or 0.2500 L).

We want to know the mass of CuSO₄·5H₂O we need to weigh on the balance.

This is a two step calculation. Let’s take a look at the expanded road map.

The first step involves using the concentration and the volume of the solution to determine how many moles of CuSO₄·5H₂O is in the solution.

Step 1: (0.2500 L) x (0.0380 M) = 0.00950 moles of CuSO₄.

Step 2: Determine how many grams of CuSO₄·5H₂O is equivalent to 0.00950 moles of CuSO₄.

We need the molar mass of 0.00950 moles of CuSO₄·5H₂O. It is 249.69 ^g/_mole.

Mass of CuSO₄·5H₂O = (0.00950 moles) x ( 249.69 ^g/_mole) = 2.37 g of CuSO₄·5H₂O

We need to weigh 2.37 g of CuSO₄·5H₂O in order to prepare a 250.0 mL solution that has a concentration of 0.0380 M CuSO₄.

Answer to Question 2:

This is a dilution question. You will need the dilution formula.
C_iV_i = C_fV_f
Let the subscript i be the original or stock solution’s concentration (C_i) and volume (V_i).
Let the subscript f be the final solution’s concentration (C_f) and volume (V_f).

Given:

1. Original or stock solution’s concentration (C_i) = 0.0380 M NaOH.
2. Final solution’s concentration (C_f) = 1.25 x 10^-3 M NaOH.
3. Final solution’s volume (V_f) = 250.0 mL.

The unknown is V_i, the volume of the NaOH stock solution needed to prepare the final solution.

Let’s solve for V_i.

V_i = C_fV_f / C_i = (1.25 x 10^-3 M) x (250.0 mL) / 0.0380 M = 8.22 mL