I need help finding Moles of sodium acetate used in dilution (mol), Moles of acetic acid solution (mol), Molarity of acetic acid in diluted solution (M). **Calculation Tables**### Table 1: Finding the number of moles of sodium acetate (NaC₂H₃O₂) in solution A| Description | Value ||--------------------------------------------------|--------|| Molarity of sodium acetate on bottle | 0.200 M || Volume of sodium acetate solution used in dilution (mL) | 5.00 mL || Volume of sodium acetate solution used in dilution (L) | 0.005 L || Moles of sodium acetate used in dilution (mol) | |### Table 2: Finding the molarity of acetic acid (HC₂H₃O₂) in solution B| Description | Value ||--------------------------------------------------|----------|| Molarity of acetic acid solution on bottle | 1.00 M || Volume of acetic acid solution used in dilution (mL) | 25.0 mL || Volume of acetic acid solution used in dilution (L) | 0.025 L || Moles of acetic acid solution (mol) | || Total volume of solution in dilution (mL) | 100.0 mL || Total volume of solution in dilution (L) | 0.1 L || Molarity of acetic acid in diluted solution (M) | |

Molarity is defined as number of moles of solute dissolved in 1 L of water. Molarity (M) = moles of solute/Volume of solution (L)Solution A [CH3COONa] = 0.200M VCH3COONa = 0.005 L Number of moles of sodium acetate (nCH3COONa) = [CH3COONa] x VCH3COONa = 0.200 M x 0.005 L = 0.001 mol Solution B [CH3COOH] = 1M VCH3COONa = 0.025L Number of moles of acetic acid(nCH3COOH) = [CH3COOH] x VCH3COOH= 1 M x 0.025 L = 0.025 mol Now total volume of solution in dilution = 0.1 L Molarity = (nCH3COOH)/VCH3COONa= 0.025mol/1L =0.025M