Problem 110 - A Complete Resource Book in Chemistry for JEE Main

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) \(100.43\)

Step-by-Step Solution

Answer

The molality of the solution is 10.43 mol/kg, option (a).

Step 1: Understanding Percent w/v

The solution contains 93% sulfuric acid, meaning 93 grams of \( \text{H}_2\text{SO}_4 \) per 100 mL of solution.

Step 2: Calculating Total Mass of Solution

The density of the solution is 1.84 g/mL. Therefore, 1 L (1000 mL) of solution has a mass of \( 1.84 \times 1000 = 1840 \) grams.

Step 3: Calculating Mass of \( \text{H}_2\text{SO}_4 \)

Using the 93% \( \text{w/v} \) concentration, 1 L of solution contains \( 93 \times 10 = 930 \) grams of sulfuric acid.

Step 4: Calculating Mass of Solvent (Water)

The mass of the water in the solution is the total mass of the solution minus the mass of sulfuric acid: \( 1840 \text{ g} - 930 \text{ g} = 910 \text{ g} \).

Step 5: Converting Solvent Mass to Kg

Since molality is the number of moles of solute per kilogram of solvent, convert the mass of water to kilograms: \( 910 \text{ g} \times \frac{1 ext{ kg}}{1000 ext{ g}} = 0.91 \text{ kg} \).

Step 6: Calculating Moles of \( \text{H}_2\text{SO}_4 \)

The molar mass of sulfuric acid \( (\text{H}_2\text{SO}_4) \) is approximately \( 98.08 \text{ g/mol} \). The number of moles of \( \text{H}_2\text{SO}_4 \) is \( \frac{930}{98.08} \approx 9.48 \text{ moles} \).

Step 7: Calculating Molality

Molality \((m)\) is calculated as the moles of solute per kilogram of solvent:\[ m = \frac{9.48 \text{ moles}}{0.91 \text{ kg}} \approx 10.43 \text{ mol/kg} \].

Key Concepts

Solution Concentration

Solution concentration refers to how much of a solute is present in a given volume of solution. It essentially tells you the strength or purity of a solution. A common way to describe solution concentration is by using percent weight/volume (% w/v), which is the amount of solute (in grams) per 100 milliliters of solution.

In the given exercise, a 93% w/v sulfuric acid solution means there are 93 grams of sulfuric acid for every 100 mL of solution. This describes how concentrated the sulfuric acid is in the solution, helping you understand its potential reactivity and strength.

• High % w/v indicates a more concentrated solution.
• Low % w/v indicates a more diluted solution.

Density

Density is a physical property defined as the mass per unit volume of a substance. It gives insight into how tightly packed the particles in a material are. When you're dealing with solutions, knowing the density allows you to relate mass and volume, which is crucial for various calculations.

In this exercise, the density of the sulfuric acid solution is given as 1.84 g/mL. This means every milliliter of this solution weighs 1.84 grams. By knowing the density, we can easily calculate the total mass of the solution when given its volume. For 1 liter of solution, which equals 1000 mL, the total mass is calculated as:
\[ \text{Mass} = 1.84 \times 1000 = 1840 \text{ grams} \]
This step is essential when transitioning from volume percent to weight calculations.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is the sum of the atomic masses of all atoms in a molecule and is crucial for converting between mass and moles.

For sulfuric acid, the formula is \( \text{H}_2\text{SO}_4 \), and its molar mass is calculated by adding the masses of 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms. This totals to approximately 98.08 g/mol, enabling us to compute the number of moles present in a given mass.

In the solution, there are 930 grams of sulfuric acid. To find how many moles this represents, we use the formula:
\[ \text{Moles} = \frac{\text{Mass of } \text{H}_2\text{SO}_4}{\text{Molar mass of } \text{H}_2\text{SO}_4} = \frac{930}{98.08} \approx 9.48 \text{ moles} \]
This step is foundational for calculating molality.

Mass Percent

Mass percent is a way of expressing a concentration as the mass of a solute divided by the total mass of the solution, multiplied by 100 to get a percentage. This gives a clear picture of the proportion of a solute within the entire solution.

• Mass percent helps identify the exact amount of solute compared to the total mass of the solution.
• Useful in complex calculations where specific proportions are essential.

In the exercise, once the total mass of sulfuric acid is known (930 grams in a 1-liter solution), you can view this as part of the entire mass, which is 1840 grams. Though the calculation focuses on molality, understanding mass percent provides additional insight into the composition of the solution. It supports the understanding of the solution's behavior in reactions, safety measures, and dilution requirements.