Problem 27 - A Complete Resource Book in Chemistry for JEE Main

Question

4 grams of an ideal gas occupies \(5.6035\) litres of volume at \(546 \mathrm{~K}\) and 2 atm pressure. What is its molecular weight? (a) 4 (b) 16 (c) 32 (d) 64

Step-by-Step Solution

Answer

The molecular weight of the gas is 16.

Step 1: Identify Known Values

We are given that 4 grams of an ideal gas occupies 5.6035 litres at 546 K and 2 atm. Let's break it down: mass = 4 grams, volume = 5.6035 litres, temperature = 546 K, pressure = 2 atm.

Step 2: Use Ideal Gas Law

The ideal gas law is given by the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature. We will use this formula to find the number of moles.

Step 3: Calculate the Number of Moles

Rearranging the ideal gas law, we have \( n = \frac{PV}{RT} \). Substituting the given values, \( R = 0.0821 \; \text{L atm/mol K} \), \( P = 2 \; \text{atm} \), \( V = 5.6035 \; \text{L} \), \( T = 546 \; \text{K} \), we get \( n = \frac{2 \times 5.6035}{0.0821 \times 546} \approx 0.25 \; \text{moles}. \)

Step 4: Find Molecular Weight

Molecular weight \( M \) is given by \( M = \frac{\text{mass in grams}}{n} \). Substituting the values, \( M = \frac{4}{0.25} = 16. \) Hence, the molecular weight is 16.

Key Concepts

Molecular Weight Calculation

Molecular weight, often referred to as molar mass, is a measure of the mass of a given substance's molecules. It is typically expressed in units of grams per mole (g/mol). To calculate molecular weight:

• Identify the mass of the substance. This can be given in grams for practical calculations.
• Determine the number of moles of the substance, which might be derived from using gas laws.
• Apply the formula: Molecular weight \( M = \frac{\text{mass in grams}}{\text{moles}} \).

In the context of an ideal gas, as in our exercise, we used the calculated moles from the gas law to deduce the molecular weight. This way, we determined that the molecular weight of the gas in question is 16 g/mol.

Gas Laws

Gas laws are simple mathematical models that describe how gases behave under various conditions of pressure, volume, and temperature. The ideal gas law is one such model represented by the equation \( PV = nRT \). Let's break down this formula:

• \( P \) stands for pressure, measured in atmospheres (atm).
• \( V \) is the volume of the gas, commonly measured in liters.
• \( n \) signifies the number of moles of the gas, which tells us how much of the substance we have.
• \( R \) is the ideal gas constant, which for our purposes is \( 0.0821 \; \text{L atm/mol K} \).
• \( T \) refers to the temperature in Kelvin (K).

The ideal gas law links these properties, allowing us to determine unknown quantities when others are known. By using these relationships, we were able to find the number of moles needed in the original exercise.

Moles Calculation

The concept of moles allows chemists to count particles in a substance by relating them to a unit that's easy to measure. Specifically, one mole represents \( 6.022 \times 10^{23} \) particles, atoms, or molecules, commonly referred to as Avogadro’s number.

In the ideal gas law, the number of moles \( n \) can be calculated using the formula \( n = \frac{PV}{RT} \), where \( P \) (pressure), \( V \) (volume), and \( T \) (temperature) are typically known or measured, and \( R \) is constant.

Using the given conditions from our exercise:

• Substituted values into the equation with \( P = 2 \; \text{atm} \), \( V = 5.6035 \; \text{L} \), \( R = 0.0821 \; \text{L atm/mol K} \), and \( T = 546 \; \text{K} \).
• Calculated \( n \) to find \( 0.25 \; \text{moles} \) of gas present.

This calculation not only facilitated the finding of molecular weight but also demonstrated one of the fundamental applications of stoichiometry in chemistry.