(3 points total) Calculate and match the volume in mL of a solution required to provide the following: 1. 2.14 g sodium chloride from a 0.270 M solution 2. 4.30 g of ethanol from a 1.50 M solution 3. 0.85 g of acetic acid from a 0.30 M solution

(3 points total) Calculate and match the volume in mL of a solution required to provide the following: 1. 2.14 g sodium chloride from a 0.270 M solution 2. 4.30 g of ethanol from a 1.50 M solution 3. 0.85 g of acetic acid from a 0.30 M solution

Answer

To find the volume of a solution required to provide a certain mass of solute, use the formula: Volume (L) = moles / Molarity. Then convert L to mL. Part 1: 2.14 g NaCl from 0.270 M - Molar mass of NaCl = 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol. - Moles of NaCl = 2.14 g / 58.44 g/mol = 0.036619 mol. - Volume in L = 0.036619 mol / 0.270 mol/L = 0.1356 L. - Volume in mL = 0.1356 L * 1000 = 135.6 mL (rounds to 136 mL). Part 2: 4.30 g ethanol (C2H5OH) from 1.50 M - Molar mass of C2H5OH = (2 * 12.01) + (6 * 1.008) + 16.00 = 46.07 g/mol. - Moles of ethanol = 4.30 g / 46.07 g/mol = 0.093336 mol. - Volume in L = 0.093336 mol / 1.50 mol/L = 0.06222 L. - Volume in mL = 0.06222 L * 1000 = 62.2 mL. Part 3: 0.85 g acetic acid (CH3COOH) from 0.30 M - Molar mass of CH3COOH = (2 * 12.01) + (4 * 1.008) + (2 * 16.00) = 60.05 g/mol. - Moles of acetic acid = 0.85 g / 60.05 g/mol = 0.014155 mol. - Volume in L = 0.014155 mol / 0.30 mol/L = 0.04718 L. - Volume in mL = 0.04718 L * 1000 = 47.18 mL (rounds to 47.2 mL, though the option provided is 47.0 mL).