Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the correct sequence of active masses in increasing order in gaseous mixture, containing one gram per litre of each of the following? 1\. \(\mathrm{NH}_{3}\) 2\. \(\mathrm{N}_{2}\) 3\. \(\mathrm{H}_{2}\) 4\. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: (a) \(3,1,4,2\) (b) \(3,4,2,1\) (c) \(2,1,4,3\) (d) \(4,2,1,3\)

Step-by-Step Solution

Answer

The correct sequence is option (d) \(4, 2, 1, 3\).

Step 1: Determine Molar Masses

First, calculate the molar mass of each gas to understand their respective number of moles per gram. The molar masses are: \( \mathrm{NH}_3 = 17 \, \text{g/mol} \), \( \mathrm{N}_2 = 28 \, \text{g/mol} \), \( \mathrm{H}_2 = 2 \, \text{g/mol} \), and \( \mathrm{O}_2 = 32 \, \text{g/mol} \).

Step 2: Calculate Active Mass (concentration in mol/L)

The active mass of each gas is the concentration in mol/L, calculated as \( \frac{\text{mass}}{\text{molar mass}} \). Given 1 gram per liter for each, the concentrations are:\ \( \frac{1}{17} \) for \( \mathrm{NH}_3 \), \( \frac{1}{28} \) for \( \mathrm{N}_2 \), \( \frac{1}{2} \) for \( \mathrm{H}_2 \), and \( \frac{1}{32} \) for \( \mathrm{O}_2 \).

Step 3: Order Concentrations

Arrange these concentrations in increasing order. \( \frac{1}{32} (\mathrm{O}_2) < \frac{1}{28} (\mathrm{N}_2) < \frac{1}{17} (\mathrm{NH}_3) < \frac{1}{2} (\mathrm{H}_2) \).

Step 4: Match to Answer Choices

Now that we have the order, compare it with the choices given. The sequence \( 4, 2, 1, 3 \) matches with option (d).

Key Concepts

Active Mass Calculation

In the realm of chemistry, particularly when dealing with gaseous mixtures, the concept of active mass plays a significant role. Active mass refers to the concentration of a substance when it is involved in a chemical reaction, especially in terms of its moles per unit volume. This concept is central when one analyzes reactions in gaseous states.
To calculate the active mass of a gas, you would typically determine its concentration in terms of moles per liter (mol/L). This involves dividing the mass of the gas by its molar mass. For example, if you have a gas present at 1 gram per liter, you can calculate its concentration using the formula: \[\text{Concentration (mol/L)} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\].
Gases in a mixture may have different active masses due to varying molar masses, directly influencing their concentration in the mixture. Remember, the higher the molar mass, the smaller the active mass for the same number of grams, affecting the order in which gases react or exert pressure in a system.

Molar Mass Determination

Molar mass is a fundamental concept in chemistry. It represents the mass of a given chemical element or compound divided by the amount of substance in moles (its molecular weight in grams per mole). Knowing how to calculate and apply it is crucial for tasks like determining the active mass or concentration of substances.
Determining the molar mass of a compound involves summing the atomic masses of all the atoms in its formula. For instance:

• For ammonia \( \mathrm{NH}_3 \), the molar mass calculation is: \[ 14 \text{(N)} + 3 \times 1 \text{(H)} = 17 \, \text{g/mol} \]
• For nitrogen gas \( \mathrm{N}_2 \), use: \[ 2 \times 14 = 28 \, \text{g/mol} \]
• For hydrogen gas \( \mathrm{H}_2 \), it's: \[ 2 \times 1 = 2 \, \text{g/mol} \]
• And for oxygen gas \( \mathrm{O}_2 \), you'll need: \[ 2 \times 16 = 32 \, \text{g/mol} \]

Understanding these calculations allows one to derive other important properties such as the concentration and helps when reacting different gases or calculating the stoichiometry in chemical reactions.

Concentration in Chemistry

Concentration plays an essential role in chemistry as it dictates how substances interact with one another, particularly in mixtures. It usually denotes the amount of a substance per defined space, often expressed in mol/L (molarity).
In gaseous mixtures, concentration is crucial for determining the behavior of gases and predicting the outcome of chemical reactions. It helps to know how much of a particular reactant is present when a reaction occurs. Concentration is calculated by dividing the number of moles by the volume in liters:

• For example, for \( 1 \text{ gram of } \mathrm{H}_2 \, \text{per liter} \), the concentration is \( \frac{1}{2} \, \text{mol/L} \), given \( \mathrm{H}_2 \) has a molar mass of \( 2 \, \text{g/mol} \).
• Similarly, for \( \mathrm{NH}_3 \), it would be \( \frac{1}{17} \, \text{mol/L} \).
• \( \mathrm{O}_2 \) yields \( \frac{1}{32} \, \text{mol/L} \), and \( \mathrm{N}_2 \) produces \( \frac{1}{28} \, \text{mol/L} \).

This concentration tells us not only about the chemical potential in reaction processes but also highlights the way gases behave under certain conditions in a mixture.