a vinegar contains acetic acid,CH3COOH.Titration of 5.00g of vinegar with 0.100M of NAOH requires 33.00cm to reach the equivalence point.a)what is the weight percentage of CH3COOH in vinegar?

The weight is 15.2%

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To determine the weight percentage of CH3COOH in vinegar, we need to calculate the number of moles of CH3COOH and the total weight of vinegar.

Step 1: Calculate the number of moles of CH3COOH:
The molar mass of CH3COOH is the sum of the atomic masses of its constituent elements (C, H, O).

Molar mass of C = 12.01 g/mol
Molar mass of H = 1.008 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of CH3COOH = (12.01 * 2) + (1.008 * 4) + 16.00 = 60.05 g/mol

To calculate the number of moles of CH3COOH, we can use the formula:

Number of moles = mass / molar mass

Mass of CH3COOH = 5.00 g

Number of moles of CH3COOH = 5.00 g / 60.05 g/mol = 0.083 moles

Step 2: Calculate the total weight of vinegar:
The total weight of vinegar is the sum of the weight of CH3COOH and the weight of the solvent (usually water).

Using the formula:

Weight percentage = (mass of solute / mass of solution) x 100

The mass of the solution is the total weight of vinegar.

Let's assume the weight of water in vinegar is negligible, which is often the case.

Total weight of vinegar = 5.00 g

Step 3: Calculate the weight percentage:

Weight percentage of CH3COOH = (mass of CH3COOH / total weight of vinegar) x 100

Weight percentage of CH3COOH = (5.00 g / 5.00 g) x 100 = 100%

Therefore, the weight percentage of CH3COOH in vinegar is 100%.

To calculate the weight percentage of CH3COOH in vinegar, you need to determine the number of moles of CH3COOH and the total mass of the vinegar.

Step 1: Calculate the number of moles of CH3COOH
Given that 0.100 M NaOH was used in the titration and 33.00 cm3 of NaOH was required to reach the equivalence point, we can calculate the number of moles of NaOH used.
Moles of NaOH = (concentration of NaOH) × (volume of NaOH in liters)
Moles of NaOH = (0.100 mol/L) × (33.00 cm3 ÷ 1000 cm3/L)
Moles of NaOH = 0.0033 mol

Since this is a 1:1 acid-base reaction, the number of moles of CH3COOH in the vinegar is also 0.0033 mol.

Step 2: Calculate the mass of CH3COOH
The molar mass of CH3COOH can be calculated based on the atomic masses of its elements:
CH3COOH = (1 × 12.01 g/mol) + (3 × 1.01 g/mol) + (1 × 16.00 g/mol) + (1 × 16.00 g/mol) = 60.05 g/mol

Mass of CH3COOH = (moles of CH3COOH) × (molar mass of CH3COOH)
Mass of CH3COOH = 0.0033 mol × 60.05 g/mol
Mass of CH3COOH = 0.1982 g

Step 3: Calculate the weight percentage of CH3COOH
Weight percentage of CH3COOH = (mass of CH3COOH ÷ mass of vinegar) × 100%
Weight percentage of CH3COOH = (0.1982 g ÷ 5.00 g) × 100%
Weight percentage of CH3COOH = 3.96% (rounded to two decimal places)

Therefore, the weight percentage of CH3COOH in vinegar is approximately 3.96%.