Question

The molality of a \(15 \%(\mathrm{w} / \mathrm{v})\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) of density \(1.1 \mathrm{~g} / \mathrm{cm}^{3}\) is nearly (a) \(1.3\) (b) \(1.6\) (c) \(1.46\) (d) \(1.86\)

Short answer

The molality of the solution is approximately 1.6. The correct option is (b).

Step-by-step solution

Understand the Problem

We need to calculate the molality (m) of a sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) solution which is given as a 15% weight/volume solution. The density of the solution is also provided as 1.1 g/cm³. Molality is defined as moles of solute per kilogram of solvent.

Determine Mass of Solution

The percentage weight/volume (w/v) of 15% means 15 grams of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is present in 100 mL of solution. The density of the solution, 1.1 g/cm³, allows us to determine the mass of the solution when the volume is known.

Calculate the Mass of the Solution

Using the density and volume, calculate the mass of the solution: \[\text{mass of solution} = \text{density} \times \text{volume} = 1.1 \times 100 = 110 \text{ grams}\]

Determine the Mass of Solvent

Subtract the mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) from the total mass of the solution to find the mass of the solvent: \[\text{mass of solvent} = 110 - 15 = 95 \text{ grams} = 0.095 \text{ kilograms}\]

Calculate the Moles of Solute

Find the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) using its molar mass (\(\mathrm{H}_{2}\mathrm{SO}_{4}\) molar mass is approximately 98 g/mol): \[\text{moles of } \mathrm{H}_{2}\mathrm{SO}_{4} = \frac{15}{98} \approx 0.153 \text{ moles}\]

Calculate Molality

Now, calculate the molality using the formula for molality: \[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.153}{0.095} = 1.611\]This value is closest to option (b), 1.6.

Key concepts

Density and Concentration

When dealing with solutions, two important parameters to understand are density and concentration. Density is the mass of a substance per unit volume. It allows us to convert between volume and mass, which is particularly useful when measuring out liquids where volume measurements are more convenient than weighing. For example, if a solution has a density of \(1.1 \text{ g/cm}^3\), it means 1 cubic centimeter (or 1 mL) of the solution weighs 1.1 grams. Understanding this concept is crucial when determining the mass of the entire solution from its volume.

Concentration, on the other hand, describes how much solute is present in a given amount of solution. The percentage weight/volume (w/v) concentration denotes how many grams of solute are present in 100 mL of solution. For instance, a 15% w/v concentration of sulfuric acid means there are 15 grams of sulfuric acid in every 100 mL of solution.

Grasping these concepts can help in various calculations, such as determining how much solute is present in a given volume or how dense a particular solution is.

Sulfuric Acid Solution

Sulfuric acid, with the chemical formula \(\mathrm{H}_{2}\mathrm{SO}_{4}\), is a heavy, syrupy, extremely strong acid. In solution form, such as in the example provided with a 15% w/v concentration, it means that in every 100 mL of the solution, there are 15 grams of \(\mathrm{H}_{2}\mathrm{SO}_{4}\).

Understanding the composition of this solution is crucial when considering chemical reactions, dilution processes, or safety procedures since sulfuric acid is highly corrosive. It reacts vigorously with water and organic material, so using such solutions requires appropriate safety equipment and handling procedures.

Due to its high density and concentration, solutions of sulfuric acid can be particularly unique to work with in laboratory settings compared to more dilute or less dense solutions.

Moles and Molar Mass

To calculate molality, knowing how to find moles from the mass of a substance is crucial. The number of moles of a substance is calculated using its molar mass. Molar mass is the mass of one mole of a compound, and for \(\mathrm{H}_{2}\mathrm{SO}_{4}\), it is approximately 98 g/mol, derived as follows:

• Hydrogen (H): 1(g/mol) \(\times 2\)
• Sulfur (S): 32(g/mol)
• Oxygen (O): 16(g/mol) \(\times 4\)

Adding these together gives you approximately 98 g/mol.

Given the mass of sulfuric acid (15 grams) within the solution, computing the moles involves dividing the solute's mass by its molar mass: \[\text{moles of } \mathrm{H}_{2}\mathrm{SO}_{4} = \frac{15}{98} \approx 0.153 \text{ moles}\].

Understanding moles and molar mass is key to converting between mass and moles, which allows deeper insight into how much of a substance is participating in a chemical reaction.

Solvent Mass Determination

Determining the mass of the solvent is an important step in calculating molality. In our example, the solution's total mass is first found using its density and volume. Here, with a density of \(1.1 \text{ g/cm}^3\) and a known solution volume of 100 mL, the total mass is determined to be \(110\) grams.

To find the mass of the solvent alone, we must subtract the mass of the solute from the total mass. With \(15\) grams of sulfuric acid in the solution, the solvent therefore (likely water, in this case) has a mass of:\[\text{mass of solvent} = 110 - 15 = 95 \text{ grams} = 0.095 \text{ kilograms}\].

This step is crucial because molality is defined as the moles of solute per kilogram of solvent, not per kilogram of solution, which distinguishes it from other concentration measures, such as molarity that uses the entire solution volume.