3.Use dimensional analysis for any/all math **Titration Problem Explanation****Problem:**A food chemist determines the concentration of acetic acid in a sample of apple cider vinegar by acid-base titration. - **Density of the sample:** 1.01 g/mL- **Titrant:** 1.002 M NaOH- **Average volume of titrant required:** 20.78 mL- **Volume of vinegar aliquot:** 25.00 mLCalculate the concentration of acetic acid in the vinegar, expressing the answer as a percent by mass.**Details to Consider:**1. **Acetic Acid Reaction:** - The acetic acid (CH₃COOH) reacts with sodium hydroxide (NaOH) in a 1:1 molar ratio.2. **Calculate Moles of NaOH:** - Use the formula: \[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (L)} \]3. **Determine Moles of Acetic Acid:** - Since the reaction is 1:1, moles of acetic acid = moles of NaOH.4. **Mass of Acetic Acid:** - Calculate the mass using: \[ \text{Mass} = \text{Moles} \times \text{Molar Mass of Acetic Acid (60.05 g/mol)} \]5. **Convert to Percent by Mass:** - Use the density to calculate the total mass of the vinegar aliquot. - Calculate percent by mass using: \[ \%\text{(w/w)} = \left(\frac{\text{Mass of Acetic Acid}}{\text{Total Mass of Vinegar}}\right) \times 100 \]This thorough explanation involves calculation steps and chemical reactions for determining the concentration of acetic acid in the vinegar sample through titration.

3. 20.78 mL of 1.002 M NaOH titrant is required to titrate 25.00 mL aliquot of vinegar sample . The density ofthe sample is 1.01 g/mL .We have to calculate the concentration of acetic acid as percent by mass in the vinegar . Answer,The acid-base chemical reaction between acetic acid and NaOH takes place in the following way - CH3COOH aq + NaOH aq → CH3COONa aq + H2O lFrom the above balanced molecular reaction, the mole ratio of acetic acid and NaOH is - CH3COOH : NaOH 1 : 1 From stoichiometric coefficientSo, number of moles of CH3COOH (nCH3COOH) = Number of moles of NaOH (nNaOH)Volume of NaOH VNaOH = 20.78 mLOr, VNaOH = 20.78 mL × 1 L1000 mL Since, 1 L = 1000 mLOr, VNaOH = 20.78 ×11000 L Or, VNaOH = 0.02078 LConcentration of NaOH CNaOH = 1.002 MSo, number of moles of NaOH nNaOH required for titration = VNaOH × CNaOHOr, nNaOH = 0.02078 L × 1.002 MOr, nNaOH = 0.02078 L × 1.002 moleL Since, molarity = MoleLOr, nnNaOH = 0.02078 × 1.002 moleOr, nnNaOH = 0.02082 mole∴ Number of moles of acetic acid present in the sample nCH3COOH = 0.02082 moleMolar mass of CH3COOH = 60.052 g/mol Thus, mass of acetic acid present in the sample mCH3COOH = nCH3COOH × Molar massOr,mCH3COOH = 0.02082 mole × 60.052 gmoleOr, mCH3COOH = 0.02082 × 60.052 gOr, mCH3COOH = 1.250 gNow, density of sample d = 1.01 g/mLVolume of aliquot of vinegar VVinegar = 25.00 mLDensity is defined as the mass of a substance per unit volume .Density = MassVolumeOr, Mass = Volume × Density So, mass of vinegar sample mVinegar = VVinegar × dOr, mVinegar = 25.00 mL × 1.01 gmLOr, mVinegar = 25.00 × 1.01 gOr, mVinegar = 25.25 gMass percent or (m/m) concentration is defined as the mass of solute dissolved in 100 grams ofsolution .∴ m/m % = Mass of solute gMass of solution g × 100 %Here, mass of solute mCH3COOH = 1.250 gMass of solution mVinegar = 25.25 g Thus, mass percent = 1.250 g25.25 g × 100 %Or, m/m = 1.25025.25 × 100 %Or, m/m = 4.9504950 %Or, m/m = 4.95 % Rounded to three significant digitsHence, the concentration of acetic acid is 4.95 % in the vinegar .