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Chapter 10: Problem 37

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, molality, mole fraction, and normality of the citric acid solution are 19.29%, 1.245 mol/kg, 0.0219, and 4.11 N, respectively.

Step by step solution

01

Determine the mass of citric acid in the solution

First, we need to find the mass of citric acid in the solution. To do this, we can use the given molarity and volume of the solution to find the moles of citric acid, then multiply by the molar mass of citric acid to find the mass. Given molarity(M) = 1.37 mol/L Density of the solution = 1.10 g/cm³ = 1.10 g/mL (1 mL = 1 cm³) Now determine the molar mass of citric acid (H3C6H5O7): = 3x1 + 6x12 + 5x1 + 7x16 = 3 + 72 + 5 + 112 = 192 g/mol Assuming the volume of solution is 1 L, we can find the mass of citric acid. mass of citric acid = molarity × volume × molar mass = 1.37 mol/L × 1000 mL × 192 g/mol = 262.944 g
02

Determine the mass of the solution

Now, using this mass, we can determine the mass of the entire solution (water + citric acid), to use as a basis for our other calculations. Mass of water in the solution = density × volume = 1.10 g/mL × 1000 mL = 1100 g Mass of the solution = mass of citric acid + mass of water = 262.944 g + 1100 g = 1362.944 g
03

Calculate the mass percent

Mass percent relates the mass of the solute(citric acid) to the mass of the solution. Mass percent = (mass of citric acid / mass of the solution) × 100 = (262.944 g / 1362.944 g) × 100 = 19.29 %
04

Calculate the molality

Molality relates the amount (moles) of solute to the mass (kg) of solvent(water). Molality = moles of citric acid / mass of water (in kg) moles of citric acid = 1.37 mol mass of water = 1100 g = 1.1 kg molality = 1.37 mol / 1.1 kg = 1.245 mol/kg
05

Calculate the mole fraction

Mole fraction is the ratio of moles of one component to the total moles of the components in the mixture. moles of water = mass of water / molar mass of water = 1100 g / 18.015 g/mol = 61.095 mol Mole fraction of citric acid = moles of citric acid / (moles of citric acid + moles of water) = 1.37 mol / (1.37 mol + 61.095 mol) = 0.0219
06

Calculate the normality

Normality is the molarity of species that produce or consume protons (H+) in an acid-base reaction. Since citric acid has three acidic protons, we need to multiply its molarity by the number of acidic protons. Normality = molarity × number of acidic protons = 1.37 mol/L × 3 = 4.11 N Therefore, the mass percent, molality, mole fraction, and normality of the citric acid solution are 19.29%, 1.245 mol/kg, 0.0219, and 4.11 N, respectively.

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