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Chapter 11: Problem 45

A \(1.37-M\) solution of citric acid $\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)$ in water has a density of 1.10 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate the mass percent, molality, mole fraction, and normality of the citric acid. Citric acid has three acidic protons.

Short Answer

Expert verified
The mass percent of citric acid in the solution is 23.91%, the molality is 1.64 m, the mole fraction is 0.0286, and the normality is 4.11 N.

Step by step solution

01

Calculate the molar mass of citric acid

First, we need to find the molar mass of citric acid: H3C6H5O7. Using the periodic table, we find the molar mass of each element and multiply it by the corresponding subscript: Molar mass of citric acid = (3 × 1.01 g/mol) + (6 × 12.01 g/mol) + (5 × 1.01 g/mol) + (7 × 16.00 g/mol) = 192.12 g/mol
02

Calculate the mass of citric acid and the mass of the solution

We are given the molarity and density of the solution. We can use these values to find the mass of citric acid and the mass of the solution in 1 L: Mass of citric acid = Molarity × Molar mass × Volume of the solution Mass of citric acid = 1.37 mol/L × 192.12 g/mol × 1 L = 263.00 g Mass of the solution = Density × Volume of the solution Mass of the solution = 1.10 g/cm³ × 1000 cm³ = 1100 g
03

Calculate the mass percent of citric acid

Now, we can use the mass of citric acid and the mass of the solution to calculate the mass percent of citric acid: Mass percent of citric acid = (mass of citric acid ÷ total mass of the solution) × 100 Mass percent of citric acid = (263.00 g ÷ 1100 g) × 100 = 23.91%
04

Calculate the molality of citric acid

To find the molality of citric acid, we need to know the moles of citric acid and the mass of the solvent in kg. Moles of citric acid = 1.37 mol/L × 1 L = 1.37 mol Mass of solvent (water) = mass of the solution - mass of citric acid = 1100 g - 263.00 g = 837.00 g = 0.837 kg Now, we can use these values to find the molality: Molality of citric acid = moles of citric acid ÷ mass of solvent (in kg) Molality of citric acid = 1.37 mol ÷ 0.837 kg = 1.64 m
05

Calculate the mole fraction of citric acid

To find the mole fraction, we need to know the total moles of solute (citric acid) and solvent (water): Moles of water = mass of water ÷ molar mass of water = 837.00 g ÷ 18.02 g/mol = 46.45 mol Now, we can find the mole fraction of citric acid: Mole fraction of citric acid = moles of citric acid ÷ total moles of solute and solvent Mole fraction of citric acid = 1.37 mol ÷ (1.37 mol + 46.45 mol) = 0.0286
06

Calculate the normality of citric acid

Citric acid has three acidic protons (3 equivalents per mole). To find the normality, we can use the molarity and the number of acidic protons: Normality of citric acid = Molarity × number of acidic protons Normality of citric acid = 1.37 mol/L × 3 = 4.11 N In conclusion, the mass percent of citric acid is 23.91%, the molality is 1.64 m, the mole fraction is 0.0286, and the normality is 4.11 N.

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