Problem 13 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The volume strength of \(8.9 \mathrm{M} \mathrm{H}_{2} \mathrm{O}_{2}\) solution calculated at \(273 \mathrm{~K}\) and 1 atm is \(.\left(\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\) (rounded off to the nearest integer)

Step-by-Step Solution

Answer

The volume strength of the solution is 100 L.

Step 1: Identify the Problem

We need to find the volume strength of a hydrogen peroxide (H₂O₂) solution. Volume strength refers to the volume of oxygen gas, in liters, liberated at NTP (Normal Temperature and Pressure i.e., 273 K and 1 atm) by 1 L of the solution.

Step 2: Understanding Volume Strength

The volume strength is based on how much oxygen gas (O₂) is produced from the decomposition of hydrogen peroxide. This chemical reaction is: \[2 ext{H}_2 ext{O}_2 ightarrow 2 ext{H}_2 ext{O} + ext{O}_2\] This shows that 2 moles of H₂O₂ yield 1 mole of O₂.

Step 3: Use the Ideal Gas Equation

To find the volume that 1 mole of a gas occupies at NTP, use the ideal gas law equation: \( PV = nRT \).For 1 mole of any gas at 273 K and 1 atm, \( V = \frac{nRT}{P} = \frac{(1 \text{ mol}) \times (0.0821 \text{ L} \text{ atm} \text{ K}^{-1} \text{ mol}^{-1}) \times (273 \text{ K})}{1 \text{ atm}} \). Thus, the volume \( V = 22.4 \text{ L} \).

Step 4: Calculate Moles of H₂O₂ in the Solution

Given that the molarity of H₂O₂ solution is 8.9 M, it means there are 8.9 moles of H₂O₂ per liter of solution. According to the decomposition reaction, 8.9 moles of H₂O₂ will yield \( \frac{8.9}{2} = 4.45 \) moles of O₂.

Step 5: Calculate Volume of O₂ Produced

The volume of O₂ formed from 4.45 moles of O₂ at NTP is \( 4.45 \text{ moles} \times 22.4 \text{ L/mol} = 99.68 \text{ L} \).

Step 6: Final Answer

The volume strength, rounded to the nearest integer, is \( \approx 100 \text{ L} \).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is an essential principle in chemistry used to relate the pressure, volume, temperature, and amount of gas. It can be expressed as \( PV = nRT \), where:

• \( P \) stands for pressure.
• \( V \) represents volume.
• \( n \) is the number of moles of the gas.
• \( R \) is the ideal gas constant, which is \( 0.0821 \, \text{L atm K}^{-1} \text{ mol}^{-1} \).
• \( T \) is the temperature measured in Kelvin.

The Ideal Gas Law is useful for determining the behavior of a gas under different conditions and is particularly relevant when working with gases at Normal Temperature and Pressure (NTP), defined as 273 K and 1 atm. Under these conditions, 1 mole of any ideal gas occupies 22.4 liters. In the context of this exercise, using the Ideal Gas Law enables us to calculate how much space a certain amount of gas will take up.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more smaller substances. In the case of hydrogen peroxide \(\text{H}_2\text{O}_2 \), its decomposition is written as:
\[ 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \]
This reaction shows that two moles of hydrogen peroxide decompose to produce two moles of water and one mole of oxygen gas,
This is significant because the volume strength of hydrogen peroxide is defined by the volume of oxygen produced, and since decomposition yields oxygen, measuring this helps us determine how concentrated a hydrogen peroxide solution is. Understanding decomposition reactions, particularly the stoichiometry involved, is vital for making predictions about the products and their quantities.

Molarity Calculation

Molarity measures the concentration of a solution, usually in moles of solute per liter of solution. In the provided exercise, the molarity of the hydrogen peroxide solution is 8.9 M, meaning there are 8.9 moles of hydrogen peroxide in each liter of solution.

• This understanding helped us predict how much oxygen would be generated during decomposition.
• For every 2 moles of \( \text{H}_2\text{O}_2 \), 1 mole of \( \text{O}_2 \) is produced, thereby using the molarity to find the exact moles of oxygen.

The formula used was \( \text{moles of } \text{O}_2 = \frac{\text{moles of } \text{H}_2\text{O}_2}{2} \), which in this case calculated as 4.45 moles of oxygen produced from a 1 liter solution. These calculations are integral in quantifying the outcomes of reactions and understanding the nature and behavior of the chemical solutions being used.