Question

For the Experiment, 10 mL of a DIVA Sciences’ White Vinegar sample will be dispensed and diluted to 100 mL in a 100-mL Volumetric Flask. Determine the range of initial concentrations.

Diluted DIVA Sciences’ White Vinegar (5%) Solution = ____________________ M
Diluted DIVA Sciences’ White Vinegar (8%) Solution = ____________________ M

Answer 1

The range of initial concentrations for the diluted DIVA Sciences’ White Vinegar solution is 0.010 M to 0.100 M.

Step-by-step explanation:

To determine the range of initial concentrations for the diluted DIVA Sciences’ White Vinegar solution, we need to know the concentration of the original white vinegar sample. The concentration of the diluted solution can be calculated using the formula:

(C1)(V1) = (C2)(V2)

where C1 is the concentration of the original solution, V1 is the volume of the original solution, C2 is the concentration of the diluted solution, and V2 is the volume of the diluted solution.

Since the original volume is 10 mL and the diluted volume is 100 mL, we can calculate the concentration range as follows:

(C1)(10 mL) = (0.100 M)(100 mL)

Solving for C1, we find the range of initial concentrations for the diluted DIVA Sciences’ White Vinegar solution to be 0.010 M to 0.100 M.

Answer 2

To find the diluted molarity of 5% and 8% acetic acid solutions after dilution, you must first calculate the molarity of the undiluted solutions and then apply the dilution equation M1V1 = M2V2. The final molarities after dilution are approximately 0.0832 M for the 5% solution and 0.1331 M for the 8% solution.

Step-by-step explanation:

To calculate the diluted concentrations of DIVA Sciences’ White Vinegar solutions when diluted from 10 mL to 100 mL, we need to make use of the concept of dilution and the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume of the diluted solution. The mass percent can be converted to molarity using the molar mass of acetic acid (CH3COOH), which is approximately 60.05 g/mol.

For a 5% vinegar solution, if we assume a density of 1.00 g/mL (as it is close to the density of water and the solution is mostly water), then 100 mL of this solution would have a mass of 100 g. Thus, 5% of this mass would be acetic acid, which is 5 g of acetic acid. Now, we divide the mass of the acetic acid by its molar mass to get the number of moles, then divide by the total volume in liters to find the molarity (M1).

M1 for 5% = (5 g / 60.05 g/mol) / 0.1 L = 0.832 M (approximately)

The dilution formula yields M2 for the 5% solution when diluted to 100 mL:

M2 for 5% = (M1 * V1) / V2 = (0.832 M * 10 mL) / 100 mL = 0.0832 M

For an 8% solution, following the same process:

M1 for 8% = (8 g / 60.05 g/mol) / 0.1 L = 1.331 M (approximately)

And the dilution formula yields M2 for the 8% solution:

M2 for 8% = (M1 * V1) / V2 = (1.331 M * 10 mL) / 100 mL = 0.1331 M