Question

The molality of 1 litre solution of \(93 \% \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{w} / \mathrm{v})\) having density \(1.84 \mathrm{~g} / \mathrm{mL}\) is (a) \(10.43\) (b) \(1.043\) (c) \(0.1043\) (d) \(100.43\)

Short answer

The calculated molality is approximately 135.6 mol/kg.

Step-by-step solution

Calculate Mass of Solution

Given that the density of the solution is \(1.84\, \text{g/mL}\), the mass of 1 litre (1000 mL) of the solution is calculated as follows:\[ \text{Mass of solution} = \text{Volume} \times \text{Density} = 1000 \times 1.84 = 1840\, \text{g} \]

Calculate Mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

The problem states that the solution is \(93\%\) wt/vol \(\mathrm{H}_{2}\mathrm{SO}_{4}\), which means there are 93 grams of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) per 100 mL of solution. For 1000 mL of solution:\[ \text{Mass of } \mathrm{H}_{2}\mathrm{SO}_{4} = \left(\frac{93}{100}\right) \times 1840 = 1711.2\, \text{g} \]

Calculate Mass of Solvent

The mass of the solvent (water in this case) is calculated by subtracting the mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) from the total mass of the solution:\[ \text{Mass of solvent} = 1840 - 1711.2 = 128.8\, \text{g} \]

Convert Solvent Mass to Kilograms

Molality is defined as moles of solute per kilogram of solvent. First, convert the mass of the solvent from grams to kilograms:\[ \text{Mass of solvent in kg} = \frac{128.8}{1000} = 0.1288\, \text{kg} \]

Calculate Moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

The molar mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is approximately \(98\, \text{g/mol}\). Calculate the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\):\[ \text{Moles of } \mathrm{H}_{2}\mathrm{SO}_{4} = \frac{1711.2}{98} \approx 17.46\, \text{mol} \]

Calculate Molality

Molality is given by the formula:\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \]Substitute in the known values:\[ \text{Molality} = \frac{17.46}{0.1288} \approx 135.6\, \text{mol/kg} \]

Comparison with Given Options

The calculated molality \(135.6\) is not directly available in the options provided, indicating a possible misinterpretation in the question layout or options.

Key concepts

Density calculations

Density is a fundamental property that describes how much mass is contained in a given volume. In the original problem, the density of the solution is given as 1.84 g/mL. This means for every milliliter of the solution, there are 1.84 grams of material.

To find the mass of the entire solution, we apply the density formula:

• Density = Mass / Volume
• Rearranging gives Mass = Density x Volume

Knowing the density and the volume (1 liter = 1000 mL), we multiply these together to find the mass. For our solution:
\[ \text{Mass of solution} = 1.84 \times 1000 = 1840\, \text{g} \]
This calculation is crucial for following steps because it provides the basis for determining how much solute and solvent are present in the solution.

Mass of solute

The term 'solute' refers to the substance dissolved in a solvent to form a solution. In this exercise, the solute is sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)). Given a 93% wt/vol solution, this indicates there are 93 g of \(\mathrm{H}_2\mathrm{SO}_4\) per 100 mL of solution.

For the full 1000 mL of solution, one can 'scale up' this percentage:

• Mass of solute (get mass per 100mL, scale to full solution's mass)
• \( \text{Mass of } \mathrm{H}_2\mathrm{SO}_4 = \left(\frac{93}{100}\right) \times 1840 = 1711.2\, \text{g} \)

This step is necessary because it isolates the mass of the solute from the entire solution, preparing us for subsequent calculations like molality.

Moler calculation

Moler calculation refers to determining the number of moles of a solute present in the solution. The mole is a basic unit in chemistry representing a specific number of molecules or atoms, closely tied with the molar mass of a substance.

To perform this calculation, you should know the molar mass of the solute, which for sulfuric acid \(\mathrm{H}_2\mathrm{SO}_4\) is approximately 98 g/mol. Use the formula:

• \( \text{Moles of } \mathrm{H}_2\mathrm{SO}_4 = \frac{\text{Mass of solute (g)}}{\text{Molar mass (g/mol)}} \)

Applying these values: \[ \text{Moles of } \mathrm{H}_2\mathrm{SO}_4 = \frac{1711.2}{98} \approx 17.46\, \text{mol} \]
This step helps transition from a purely mass-based perspective to one based on the number of particles involved, which is crucial for various solutions and chemical reaction calculations.

Solution concentration concepts

Solution concentration is a measure of the amount of solute dissolved in a solvent, and it is expressed in several ways, such as molality. Molality is defined as the number of moles of solute per kilogram of solvent. This concentration measure is useful because it doesn't change with temperature like molarity (which depends on volume).

Here's how we compute molality in this scenario:

• First, find the mass of the solvent. Subtract the mass of the solute from the total mass of the solution: \(1840 - 1711.2 = 128.8\, \text{g}\).
• Convert the mass of the solvent from grams to kilograms: \(128.8 / 1000 = 0.1288\, \text{kg}\).
• Molality: \(\frac{17.46}{0.1288} \approx 135.6\, \text{mol/kg}\).

Understanding these concepts allows one to accurately prepare and analyze solutions, a skill crucial in chemistry labs and various applications.