Question

An aqueous solution is \(34.0 \% \mathrm{H}_{3} \mathrm{PO}_{4}\) by mass and has a density of \(1.209 \mathrm{g} / \mathrm{mL}\). What are the molarity and molality of this solution?

Short answer

By following the above calculations based on the definitions of molarity and molality, it is possible to solve this exercise. Molarity and molality are important concentrations units in chemistry, and give insights into the properties of a solution, such as boiling point, freezing point, and osmotic pressure.

Step-by-step solution

Determine mass of phosphoric acid and solution

Based on the problem, 34.0% of the solution by mass is \( \mathrm{H}_{3} \mathrm{PO}_{4}\). Therefore, in a 100 g sample of the solution, there are 34 g of \( \mathrm{H}_{3} \mathrm{PO}_{4}\). The mass of the solution (100 g) can be converted into volume using the provided density \((1.209 \mathrm{g} / \mathrm{mL})\). The volume V is calculated by dividing the mass by the density: \(V = 100 \mathrm{g}/(1.209 \mathrm{g/mL})\).

Calculate moles of phosphoric acid

The moles of solute are next to be calculated. The molar mass of phosphoric acid \( \mathrm{H}_{3} \mathrm{PO}_{4}\) is \(98.00 \mathrm{g/mol}\). Next, compute the number of moles n of phosphoric acid in the 34 g sample, \(n = 34 \mathrm{g}/(98.00 \mathrm{g/mol})\).

Calculate molarity

Molarity (M) is determined as the number of moles of the solute divided by the volume of the solution in liters. With the volume V already calculated to be represented in L and the number of moles n, the molarity is given by: \(M = n/V\).

Calculate molality

Molality (m) is given by the number of moles of the solute divided by the mass of the solvent in kilograms. The mass of water in the 100 g solution is \(100 \mathrm{g} - 34 \mathrm{g}\). Converting this to kilograms allows us to calculate the molality as: \(m = n/mass_{water}\)

Key concepts

Phosphoric Acid Solution

Phosphoric acid, with the chemical formula \( \mathrm{H}_{3} \mathrm{PO}_{4} \), is a type of inorganic acid commonly used in industries such as fertilizers and cleaning products. In chemistry, it often appears in solutions for various reactions and experiments.
When preparing a phosphoric acid solution, especially for an aqueous solution, one important parameter is the concentration of the acid. This is typically measured in terms of percentage by mass, which tells us how much of the total solution consists of phosphoric acid. For instance, a 34% by mass solution indicates that 34% of the solution's mass is due to phosphoric acid, while the rest is primarily the solvent, like water.
Understanding the mass percentage of phosphoric acid helps when calculating other properties, like molarity and molality, which define the solution's strength and its suitability for specific purposes. When entering the lab or dealing with theoretical calculations, grasping these percentage concepts is crucial for accuracy and safety.

Aqueous Solution Density

Density in the context of an aqueous solution refers to the mass of the solution per unit volume, usually expressed in \( \mathrm{g/mL} \). In our current exercise, the density is given as \( 1.209 \mathrm{g/mL} \). This value is important because it helps convert between the mass and volume of the solution.
To find the volume of a sample with known mass, you divide the mass by the density. For example, if you have 100 g of a solution, the volume is calculated as \( \frac{100 \mathrm{g}}{1.209 \mathrm{g/mL}} \). Converting the mass to volume is a crucial step in further calculations, such as determining molarity, where you need the solution's volume in liters.
An appreciation for an aqueous solution's density provides insight into its composition and behavior. It not only aids in calculations but also informs decisions regarding handling and storage, as denser solutions might behave differently than expected.

Percentage by Mass

Percentage by mass is a measure that indicates how much solute is present in a solution relative to the total mass of the solution. It is calculated using the formula:

• \( \text{Percentage by mass} = \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100 \)\%

In this problem, 34% indicates that for every 100 g of the solution, 34 g are phosphoric acid, with the remaining 66 g being solvent, primarily water.
Understanding the percentage by mass is key for converting this information when calculating molarity and molality, which instead use moles and different reference quantities. Always remember, the mass percentage provides a broad view of concentration that's easily understood and translated into practical terms for chemical work.
Grasping percentage by mass is vital for accurately making or adjusting solutions in a laboratory setting, ensuring the intended chemical reactions or processes occur as desired.