I need help with my homwork I have done procedure ABut I am not sure how to do Procedure B and C **Lab Data and Calculations****Weight and Density Measurements:**- **Weight of Water needed:** 45g- **Weight of Salt needed:** 5g- **Weight of beaker:** 78.31g- **Weight of beaker & 10.0 ml of salt solution:** 88.29g- **Weight of 10.0 ml salt solution:** 9.98g- **Density of 10% salt solution:** 0.9989 g/ml- **Reference Value for density:** 1.07 g/ml- **% Error Calculation:** 6.73%*Note: Calculations for density and error percentage are detailed on the right margin of the document.***Procedure B:**- **Mass of CuSO₄·5H₂O needed (show calculations):** __________**Procedure C:**- **Volume of 1.00 M CuSO₄ needed for 50 ml of 0.5 M CuSO₄ (show calculations):** __________- **Volume of 0.5 M CuSO₄ needed for 50 ml of 0.25 M CuSO₄ (show calculations):** __________- **Volume of 0.25 M CuSO₄ needed for 50 ml of 0.1 M CuSO₄ (show calculations):** __________**Absorbance Data:**| **Concentration (M)** | **Absorbance** ||-----------------------|----------------|| 1.00 M | 0.879 || 0.50 M | 0.452 || 0.25 M | 0.221 || 0.10 M | 0.089 |*This table shows the absorbance values for different concentrations of CuSO₄ solutions.* Ensure all calculations are shown in the students' lab reports for full credit. **Procedure A: Preparation of 50 g of a 10.0% Salt Solution**1. Calculate the amount of salt and water needed.2. Weigh the calculated amount of salt onto weighing paper to protect the balance pan.3. Weigh the calculated amount of water in a beaker.4. Add the weighed amount of salt and stir until the salt dissolves.5. Pipet 10.0 ml of the salt solution into another weighed beaker and determine the weight of the salt solution.6. Determine the density of your salt solution using the formula (Density=mass/volume).7. Find the reference value of a 10% salt solution and calculate the % error.**Procedure B: Preparation of 50.0 ml of 1.0 M CuSO₄**1. Calculate the amount of CuSO₄ needed to make 50.0 ml of a 1.0 M solution. The molecular weight of CuSO₄·5H₂O is 250 g/mole.2. Tare a small beaker and do not remove it from the balance.3. Slowly pour CuSO₄ into the beaker until the balance reads the correct weight.4. Dissolve the weighed CuSO₄ in 25-30 ml of distilled water.5. Transfer to a volumetric flask through a funnel. **If you spill any, you must start again.** Rinse the funnel into the flask with a stream of distilled water.6. Add distilled water to the mark. Add the last few drops carefully with a dropper.7. Use this **standard solution** to prepare more dilute solutions of CuSO₄ in Procedure C.**Procedure C: Preparation of More Dilute Solutions of CuSO₄**1. Calculate the size of the aliquot needed to prepare 50.0 ml of 0.50 M CuSO₄ solution from your standard solution.2. Pour a small amount of your standard solution into a small beaker and pipet the calculated aliquot from the beaker to a clean 50 ml volumetric flask. Dilute to volume.3. Stopper the flask, invert and swirl.4. Repeat the above steps to prepare 0.25 M and 0.10 M CuSO₄ solutions.5. Measure the absorbance of all four solutions that you have prepared.**Procedure D:**On the graph provided in

Molarity is used to denote the concentration of a solution. Molarity is the number of moles of solute present per liter of solvent. Given that CuSO4(5H2O) is present. Molarity of the solution = 1.0M Volume of the solution = 50.0 mL Molar mass of CuSO4(5H2O) = 250 grams Substituting the values in the molarity formulae to determine mass as follows: Molarity = Number of moles of solute1 litre of solventMolarity = Mass of soluteMolar mass of solute×1000Volume in mLMass of Solute = Molarity×Molar mass of solute×Volume in mL1000Mass of Solute =1.0 M×250 grams×50.0 mL1000 mLMass of Solute =12.5 grams Therefore, 12.5 grams of CuSO4(5H2O) should be dissolved to obtain 1.0 M in 50.0 mL of water .Molarity of CuSO4(5H2O) Solution present = 1.0 M According to dilution law: M1×V1 = M2 × V2 Where: M1 = Molarity of the stock present = 1.0M; V1 = Volume of stock needed M2 = Molarity of required solution to be prepared; V1 = Volume of required solution 50.0 mL of 0.5 M from 1.0 M Required Molarity (M2) = 0.5 M Volume of required solution (V1) = 50.0 mL Substituting the values in the Dilution law equation to determine volume of stock needed: M1×V1 = M2 × V2V1 = M2 × V2M1V1 = 0.5 M×50.0 mL1.0 MV1 = 25 mL Therefore, 25 mL of 1.0 M solution to be transferred into a 50 mL volumetric flask and then make up to the mark with distilled water to get 0.5 M solution. 1. 50.0 mL of 0.25 M from 1.0 M Required Molarity (M2) = 0.25 M Volume of required solution (V1) = 50.0 mL Substituting the values in the Dilution law equation to determine volume of stock needed: M1×V1 = M2 × V2V1 = M2 × V2M1V1 = 0.25 M×50.0 mL1.0 MV1 = 12.5 mL Therefore, 12.5 mL of 1.0 M solution to be transferred into a 50 mL volumetric flask and then make up to the mark with distilled water to get 0.25 M solution. 3. 50.0 mL of 0.5 M from 1.0 M Required Molarity (M2) = 0.1 M Volume of required solution (V1) = 50.0 mL Substituting the values in the Dilution law equation to determine volume of stock needed: M1×V1 = M2 × V2V1 = M2 × V2M1V1 = 0.1 M×50.0 mL1.0 MV1 = 5 mL Therefore, 5 mL of 1.0 M solution to be transferred into a 50 mL volumetric flask and then make up to the mark with distilled water to get 0.1 M solution.