Question
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
Solution
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(222 Votos)
Yasmin
Mestre · Tutor por 5 anos
Resposta
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:C1V1 = C2V2Where:C1 = initial concentration (1.25 N)V1 = initial volume (250 mL)C2 = final concentration (0.5 N)V2 = final volume (unknown)Rearranging the formula to solve for V2:V2 = (C1 * V1) / C2Substituting the given values:V2 = (1.25 N * 250 mL) / 0.5 NV2 = 625 mLTherefore, approximately 625 mL of water must be added to 250 mL of a 1.25 N solution to make it 0.5 N.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.Given:Molality of H2O2 = 0.610 mThe mole fraction of H2O2 can be calculated using the formula:Mole fraction of H2O2 = (Molality of H2O2) / (Molality of H2O2 + Molality of water)Since the molality of 1 m (assuming the density of water is 1 g/mL), the mole fraction of H2O2 is:Mole fraction of H2O2 = 0.610 / (0.610 + 1)Mole fraction of H2O2 ≈ 0.377Therefore, the mole fraction of hydrogen peroxide in the 0.610 m solution is approximately 0.377.51. 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.Given:Molality of citric acid = 0.688 mTo 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.Molarity of citric acid = Molality of citric acid * Density of solutionMolarity of citric acid = 0.688 m * 1.049 g/mLMolarity of citric acid ≈ 0.721 MTherefore, the molar concentration of citric acid in the 0.688 m solution is approximately 0.721 M.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.Given:Molality of acetic acid = MThe molal concentration of acetic acid can be calculated using the formula:Molal concentration of acetic acid = Molality of acetic acid * Density of solutionSubstituting the given values:Molal concentration of acetic acid = M * 1.004 g/mLTherefore, the molal concentration of acetic acid in the solution with a density of 1.004 g/mL is M * 1.004 g/mL.