Problem 118 - A Complete Resource Book in Chemistry for JEE Main

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) \(1.86\)

Step-by-Step Solution

Answer

The molality is approximately 1.6, so the correct answer is (b).

Step 1: Understanding the Given Information

We are given a 15% (w/v) solution of \( \mathrm{H}_2\mathrm{SO}_4 \) with a density of 1.1 g/cm³. This means there are 15 grams of \( \mathrm{H}_2\mathrm{SO}_4 \) in 100 mL of solution. Molality is defined as the moles of solute per kilogram of solvent.

Step 2: Calculate Moles of Solute

The molecular weight of \( \mathrm{H}_2\mathrm{SO}_4 \) is \( 2(1) + 32 + 4(16) = 98 \) g/mol. The moles of \( \mathrm{H}_2\mathrm{SO}_4 \) in 15 grams is calculated as follows: \( \frac{15}{98} = 0.153 \) moles.

Step 3: Calculate Mass of Solution

Since the density is 1.1 g/cm³, 100 mL of solution has a mass of \( 100 \times 1.1 = 110 \) g.

Step 4: Calculate Mass of Solvent

The mass of the solvent is the mass of the solution minus the mass of the solute. So, \( 110 - 15 = 95 \) g or \( 0.095 \) kg.

Step 5: Calculate Molality

Molality (m) is given by \( \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.153}{0.095} = 1.61 \).

Step 6: Select the Correct Answer

The calculated molality of the solution is 1.61 mol/kg. The answer closest to this value is (b) 1.6.

Key Concepts

Solution Concentration

Solution concentration indicates the amount of solute present in a solution, compared to the amount of solvent or the total solution. For a 15% (w/v) sulfuric acid solution, this means there are 15 grams of solute for every 100 mL of solution. Concentration can also be expressed in different terms like molarity or molality, each indicating concentration slightly differently. Understanding the concentration concept enables us to assess how much solute is in a given volume or mass of the solution, which is crucial for diverse chemical fields from industrial applications to biochemical processes.

Molecular Weight Determination

Molecular weight, sometimes referred to as "molecular mass," pertains to the mass of a given molecule. For sulfuric acid, \( \mathrm{H}_2\mathrm{SO}_4 \), the molecular weight is calculated by summing the atomic masses of all the atoms in the molecule. This involves:

• Hydrogen (H), with an atomic mass of approximately 1 g/mol, contributing to 2 g/mol since there are two hydrogen atoms.
• Sulfur (S) with its atomic mass of 32 g/mol.
• Oxygen (O) with an atomic mass of 16 g/mol, contributing to 64 g/mol from four oxygen atoms.

Together, these yield a molecular weight of 98 g/mol for \( \mathrm{H}_2\mathrm{SO}_4 \). Knowing the molecular weight is essential for calculating how many moles of a substance you have based on its mass.

Density and Mass Calculations

Density is a property that relates mass to volume. It tells you how much mass is contained in a given volume of a substance and is usually expressed in g/cm³. For the sulfuric acid solution with a density of 1.1 g/cm³, each cubic centimeter of solution weighs 1.1 grams.
To find the mass of 100 mL of this solution:

• Multiply the volume (100 mL) by the density (1.1 g/cm³), resulting in a mass of 110 grams.

Understanding these calculations enables precise measurement and conversion of volumes to mass, which is crucial when preparing solutions for any chemical reaction or experiment.

Molality Definition

Molality is a measure of solution concentration expressed as the number of moles of solute per kilogram of solvent. Unlike molarity, molality is independent of temperature because it is based on mass, not volume, making it especially useful in scenarios where temperature changes are expected.
For example, in the given exercise, molality is calculated using the formula:

• Molality (\(m\)) = \( \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
• In our case, it is \( \frac{0.153 \text{ moles}}{0.095 \text{ kg}} = 1.61 \text{ mol/kg} \).

This unit is particularly valuable in various scientific fields, including chemistry and physics, for analyzing the effects of solutes on the properties of solvents.

Steps in Problem Solving

Solving a chemistry problem often involves a series of logical steps. By breaking down the problem into manageable parts, you can systematically approach complex calculations:

• Identify what is given and what you need to find. For instance, the problem provided the percentage concentration and the density.
• Determine relevant formulas, like calculating molecular weight or using density to find mass.
• Carry out step-by-step calculations for clarity and accuracy, such as calculating moles from molecular weight or converting mass to the appropriate units.
• Always cross-check your work, and match your final answer with the possible options given in your problem.

These steps help ensure accuracy and efficiency in problem-solving, equipping you with a structured approach to tackling homework questions and real-world scenarios.