d) 3 your and 1.25 N solution to make it 0.5 OON. 1.83g/mL 49. Calculate the approxignate volume of water that must be added to 250 mL 50. An antiseptic solution contains hydrogen peroxide, H_(2)O_(2) in water. The solution 0.610 m H_(2)O_(2) What is the mole fraction of hydrogen peroxide? 51. Citric acid, H_(3)C_(4)H_(5)O_(7) occurs in plants. Lemons contain 5% to 8% citric acid by mass. The acid is added to beverages and candy. An aqueous solution is 0.688 m citric acid, The density is 1.049g/mL What is the molar concentration? is 1.004g/mL What is the molal concentration of acetic acid? is M acetic acid, HC_(2)H_(3)O_(2) The density of the vinegar

49. To calculate the approximate volume of water that must be added to 250 mL of a 1.25 N solution to make it 0.5 N, we can use the dilution formula:<br /><br />C1V1 = C2V2<br /><br />Where:<br />C1 = initial concentration (1.25 N)<br />V1 = initial volume (250 mL)<br />C2 = final concentration (0.5 N)<br />V2 = final volume (unknown)<br /><br />Rearranging the formula to solve for V2:<br /><br />V2 = (C1 * V1) / C2<br /><br />Substituting the given values:<br /><br />V2 = (1.25 N * 250 mL) / 0.5 N<br />V2 = 625 mL<br /><br />Therefore, approximately 625 mL of water must be added to 250 mL of a 1.25 N solution to make it 0.5 N.<br /><br />50. To calculate the mole fraction of hydrogen peroxide (H2O2) in a 0.610 m solution, we need to know the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.<br /><br />Given:<br />Molality of H2O2 = 0.610 m<br /><br />The mole fraction of H2O2 can be calculated using the formula:<br /><br />Mole fraction of H2O2 = (Molality of H2O2) / (Molality of H2O2 + Molality of water)<br /><br />Since the molality of 1 m (assuming the density of water is 1 g/mL), the mole fraction of H2O2 is:<br /><br />Mole fraction of H2O2 = 0.610 / (0.610 + 1)<br />Mole fraction of H2O2 ≈ 0.377<br /><br />Therefore, the mole fraction of hydrogen peroxide in the 0.610 m solution is approximately 0.377.<br /><br />51. To calculate the molar concentration of citric acid (H3C4H5O7) in a 0.688 m solution, we need to know the molarity of the solution. Molarity is defined as the number of moles of solute per liter of solution.<br /><br />Given:<br />Molality of citric acid = 0.688 m<br /><br />To convert molality to molarity, we need to know the density of the solution. Given that the density is 1.049 g/mL, we can assume the density of water is 1 g/mL.<br /><br />Molarity of citric acid = Molality of citric acid * Density of solution<br />Molarity of citric acid = 0.688 m * 1.049 g/mL<br />Molarity of citric acid ≈ 0.721 M<br /><br />Therefore, the molar concentration of citric acid in the 0.688 m solution is approximately 0.721 M.<br /><br />To calculate the molal concentration of acetic acid (HC2H3O2) in a solution with a density of 1.004 g/mL, we need to know the molality of the solution.<br /><br />Given:<br />Molality of acetic acid = M<br /><br />The molal concentration of acetic acid can be calculated using the formula:<br /><br />Molal concentration of acetic acid = Molality of acetic acid * Density of solution<br /><br />Substituting the given values:<br /><br />Molal concentration of acetic acid = M * 1.004 g/mL<br /><br />Therefore, the molal concentration of acetic acid in the solution with a density of 1.004 g/mL is M * 1.004 g/mL.