Final answer:
To calculate the mass percent, molality, mole fraction, and normality of citric acid in a 0.557 M solution, we first determine the mass percent to be 10.71%.
Step-by-step explanation:
To calculate the mass percent of citric acid in the solution, we first need to calculate the mass of citric acid in the solution. The density of the solution is given as 1.04 g/mL, which means that 1 mL of the solution has a mass of 1.04 g. So, for a 0.557 M solution, the mass of citric acid in 1 L of solution is 0.557 M x 192.13 g/mol = 107.11 g.
The mass percent of citric acid is then (107.11 g/1000 g) x 100% = 10.71%.
The molality of a solution is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is citric acid and the solvent is water. We can calculate the molality using the following formula: molality = (moles of solute) / (mass of solvent in kg).
Since the mass of citric acid is 107.11 g and the density of the solution is 1.04 g/mL, the mass of water in 1 L of solution can be calculated as 1000 g - 107.11 g = 892.89 g. Converting this to kg gives us 0.89289 kg.
Using (moles of solute) = (mass of solute / molar mass of solute), we can calculate the moles of citric acid as 107.11 g / 192.13 g/mol = 0.557 M.
Finally, the mole fraction of citric acid can be calculated as (moles of citric acid) / (moles of citric acid + moles of water) = 0.557 / (0.557 + 55.59) = 0.00995.
The normality of citric acid can be calculated based on the number of acidic protons it has. Citric acid has three acidic protons, so the normality is three times the molarity, which is 1.671 N.