Calculate the density of distilled white vinegar. Show your all work.

Density=mass/volume = 30.05g/30.00mL=1.002g/mL

3. The concentration 5% (w/v) means that 5 g of acetic acid (solute) are present in 100 mL of the solution. Using the density you determined in the previous problem, determine the mass of water (solvent) in 100 mL of the solution. Follow the guided steps below.

a. Determine the mass of 100. mL of the solution by using the density. Express your answer in three significant figures.

If I did this right I think the answer to this one is:

100mL*1.002g/mL= 100 g (mass of solution)

b. The mass of the solution should be the sum of the mass of the solvent and the mass of the solute. Now determine the mass of the solvent.

100g of solution + ____ + 5 g solute=95 g of solvent.

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Part C: Determination of Ka of Acetic Acid

4. The concentration of distilled white vinegar is written as 5% (w/v) of acetic acid (CH3COOH). Convert the concentration in the unit of molarity. Express your answer in four significant figures. Show your work here.

5. Calculate the concentration of hydronium ions [H3O+] from the measured pH values. Express your answer in two significant figures. Show your all work.

Using the initial concentration of CH3COOH (from Problem 4) and the equilibrium concentration of H3O+ calculated (from Problem 5), complete the reaction table for vinegar. Then, calculate the acidity constant. Show your work.

Reaction Table: CH3COOH (aq) + H2O(l) ?CH3COO?(aq)+ H3O+(aq)

Using the concentrations at equilibrium, the acidity constant (Ka) is calculated as

7. The acidity constant of acetic acid is 1.8 ? 10-5. Calculate the % error of the measured Ka value.

Please help I am so confussed. Thank you in advance.