Question

A solution of phosphoric acid was made by dissolving \(10.0 \mathrm{~g}\) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in \(100.0 \mathrm{~mL}\) water. The resulting volume was \(104 \mathrm{~mL}\). Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\).

Short answer

The density of the phosphoric acid solution is \(1.057 \mathrm{~g/mL}\). The mole fractions for \(H_{3}PO_{4}\) and \(H_{2}O\) are 0.0180 and 0.982, respectively. The molarity of the solution is \(0.980 \mathrm{~mol/L}\), and its molality is \(1.02 \mathrm{~mol/kg}\).

Step-by-step solution

Find the number of moles of phosphoric acid and water

To find the number of moles of phosphoric acid, we can use the formula: Number of moles = mass / molar_mass The molecular weight of phosphoric acid (H3PO4) = 3 * H + P + 4 * O = (3 * 1.01) + 30.97 + (4 * 16) = 98.0 g/mol. Now, we can find the number of moles of phosphoric acid: Number of moles of H3PO4 = 10.0 g / 98.0 g/mol = 0.102 mol For water, we have 100 mL given and knowing the density of water to be 1 g/mL, we can find the mass of water used. mass of water = density * volume = 1.00 g/mL * 100 mL = 100 g Number of moles of water = mass / molar_mass The molecular weight of water (H2O) = 2 * H + O = (2 * 1.01) + 16 = 18.02 g/mol. Number of moles of H2O = 100 g / 18.02 g/mol = 5.55 mol

Calculate the density of the solution

To find the density, we can use the formula: density = total_mass / total_volume total_mass = mass of H3PO4 + mass of water = 10 g + 100 g = 110 g total volume = 104 mL density = 110 g / 104 mL = 1.057 g/mL

Calculate the mole fraction

To find the mole fraction, use the formula: mole_fraction = moles of component / total moles The total moles of the solution = moles of H3PO4 + moles of H2O = 0.102 mol + 5.55 mol = 5.652 mol Mole fraction of H3PO4 = 0.102 mol / 5.652 mol = 0.0180 Mole fraction of H2O = 5.55 mol / 5.652 mol = 0.982

Calculate the molarity

To find the molarity, use the formula: molarity = moles of solute / volume of solution (in L) First convert the volume of the solution to liters: volume of solution = 104 mL * (1 L / 1000 mL) = 0.104 L molarity = 0.102 mol / 0.104 L = 0.980 mol/L

Calculate the molality

To find the molality, use the formula: molality = moles of solute / mass of solvent (in kg) First convert the mass of water (solvent) to kilograms: mass of water = 100 g * (1 kg / 1000 g) = 0.100 kg molality = 0.102 mol / 0.100 kg = 1.02 mol/kg The final results are: - Density of the solution = 1.057 g/mL - Mole fraction of H3PO4 = 0.0180, and mole fraction of H2O = 0.982 - Molarity of the solution = 0.980 mol/L - Molality of the solution = 1.02 mol/kg

Key concepts

Understanding Solution Density

When we discuss solution density, we are referring to the mass of the solution per unit volume. It’s a fundamental property that helps characterize solutions in chemistry. In our exercise, the density of a phosphoric acid solution is calculated by dividing the total mass of the solution by its total volume.

It is essential to realize that the solution's density can differ from the density of the pure solvent due to the addition of solute particles. This knowledge is not only critical for academic purposes but also practical in real-world applications, like preparing chemical solutions with precise concentrations in a lab.

Mole Fraction Fundamentals

Mole fraction is a way of expressing the concentration of a component in a mixture. It is the ratio of the number of moles of a single component to the total number of moles in the mixture. Students should understand that the mole fraction is dimensionless, meaning it has no units, and is useful when dealing with changes in physical properties due to composition, such as boiling point elevation or freezing point depression.

For clarity, ensuring students can visualize that a mole fraction of 0.0180 for phosphoric acid implies that out of every 1 mole of the solution, 0.0180 moles consist of phosphoric acid. By understanding this concept, they can grasp the relative amount of each substance in a mixture.

Deciphering Molarity

Molarity is a commonly used concentration unit in chemistry, representing the number of moles of solute per liter of solution. Chalking out the distinction between a solute and the solution is important for students—it helps them realize that the volume considered in molarity is that of the entire solution, not just the solvent. This unit is incredibly versatile and is often applied in calculations involving stoichiometry and dilutions.

Through examples like our exercise, students can learn to convert volumes to liters and to work with measurements effectively, an essential skill for both chemistry problems and laboratory work.

Demystifying Molality

Molality, sometimes less familiar to students than molarity, is another unit of concentration, defined as the moles of solute per kilogram of solvent. One key insight students should note is that molality is independent of temperature since mass remains unchanged with temperature fluctuations, unlike volume.

Acquainting students with molality is crucial for understanding colligative properties that depend on the number of particles dissolved in a solvent, such as boiling point elevation and freezing point depression. The exercise provided demonstrates how to convert mass to kilograms and calculate molality, a practical technique for problem-solving in various chemical contexts.