Problem 53 - A Complete Resource Book in Chemistry for JEE Main

Question

\(0.5 \mathrm{M}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is diluted from 1 litre to 10 litre, normaliy of the resulting solution is (a) \(1 \mathrm{~N}\) (b) \(10 \mathrm{~N}\) (c) \(11 \mathrm{~N}\) (d) \(0.1 \mathrm{~N}\)

Step-by-Step Solution

Answer

The normality of the resulting solution is 0.1 N, option (d).

Step 1: Understand Molarity and Normality

Molarity (M) is defined as the number of moles of solute per liter of solution. Normality (N) is the number of equivalents of solute per liter of solution. For sulfuric acid, which has two replaceable hydrogen ions, the normality is twice the molarity. Thus, initial normality = 2 x 0.5 M = 1 N.

Step 2: Use the Dilution Formula

The dilution formula is used to calculate the normality of a solution after it has been diluted: \( N_1 \times V_1 = N_2 \times V_2 \). Here, \(N_1 = 1\, \text{N}\) (initial normality), \(V_1 = 1\, \text{liter}\), \(V_2 = 10\, \text{liters}\) (final volume). We need to find \(N_2\), the normality after dilution.

Step 3: Apply Values to the Dilution Formula

Substitute the known values into the formula: \(1 \times 1 = N_2 \times 10\). Solve for \(N_2\) by dividing both sides by 10: \(N_2 = \frac{1}{10} = 0.1 \text{N}\).

Step 4: Compare with Answer Choices

The calculated normality after dilution is 0.1 N. Compare this with the provided answer choices. The correct option corresponds to \(0.1 \mathrm{~N}\), which is option (d).

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It's defined as the number of moles of solute, which is the substance being dissolved, per liter of solution. This measurement helps chemists understand how saturated a solution is with its solute.
For example, when we say a solution is 0.5 M, it means there are 0.5 moles of solute in every liter of solution. In the original problem, we see sulfuric acid used as the solute. If a student measures a 0.5 M concentration, they're handling a fairly dilute solution.
This unit of concentration provides a straightforward means to calculate how many moles are present in varying volumes, especially when it's time to dilute solutions.

Normality

Normality is closely related to molarity but takes a step further by considering the reactivity of solutes in solution. It measures the equivalent concentration of a solution. This is particularly useful in chemistry for reactions where the capacity of a solute, like acids and bases, to react is a key consideration.

• Molarity focuses on the number of moles.
• Normality, on the other hand, accounts for equivalents, which are based on a compound's role in reactions.

In the case of sulfuric acid ( H_2SO_4 ), it can release two hydrogen ions, which means its normality is twice its molarity. That's why for a 0.5 M sulfuric acid solution, the normality would be 1 N. This makes normality particularly useful for titrations and other analytical chemistry applications.

Dilution Formula

The dilution formula is a handy formula in chemistry, used to find out how a solution's concentration changes when the solution is diluted with more solvent. The formula is expressed as: \( N_1 \times V_1 = N_2 \times V_2 \), where:

• \(N_1\) is the initial normality of the solution
• \(V_1\) is the initial volume of the solution
• \(N_2\) is the final normality after dilution
• \(V_2\) is the final volume of the solution

This relationship tells us that when you dilute a solution, its concentration decreases, while the total moles or equivalents stay constant. By solving for \(N_2\), you can find the new normality after a known volume change.
For instance, in our initial exercise, when 1 liter of 1 N solution of sulfuric acid is diluted to 10 liters, the resulting normality becomes 0.1 N. This helps determine how the solution will behave in various chemical reactions.