Problem 23 - A Complete Resource Book in Chemistry for JEE Main

Question

A gas diffuses four times as quickly as oxygen. The molar weight of gas is (a) 2 (b) 4 (c) 8 (d) 16

Step-by-Step Solution

Answer

The molar weight of the gas is 2 g/mol.

Step 1: Understand Graham's Law of Diffusion

Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. So, if a gas diffuses four times as quickly as oxygen, let the molar mass of oxygen be \( M_{O_2} = 32\, \text{g/mol} \).

Step 2: Set up the Graham's Law Equation

We use the law to set up the equation: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r_1 \) is the rate of the unknown gas, \( r_2 \) is the rate of oxygen, \( M_1 \) is the molar mass of oxygen, and \( M_2 \) is the molar mass of the unknown gas.

Step 3: Substitute the Given Values Into the Equation

Since the gas diffuses four times as quickly as oxygen, \( r_1 = 4r_2 \). Substitute the values into the equation: \( \frac{4r_2}{r_2} = \sqrt{\frac{32}{M_2}} \). This simplifies to \( 4 = \sqrt{\frac{32}{M_2}} \).

Step 4: Solve for the Molar Mass of the Gas

Square both sides of the equation to remove the square root: \( 16 = \frac{32}{M_2} \). Now, solve for \( M_2 \) by multiplying both sides by \( M_2 \) and then dividing by 16: \( M_2 = \frac{32}{16} = 2 \).

Step 5: Verify the Solution

Finally, we verify by checking that the conditions satisfy Graham's Law. With \( M_2 = 2 \), the diffusion rate relation \( \frac{r_1}{r_2} = 4 \) holds true because \( 4 = \sqrt{\frac{32}{2}} \). Thus, the calculations are consistent.

Key Concepts

Rate of Diffusion

The rate of diffusion refers to how quickly a gas spreads out or mixes with another gas. This can be influenced by several factors, such as temperature, pressure, and the nature of the gases involved. However, one of the most significant factors is the molar mass of the gas. Essentially, lighter gases tend to diffuse faster than heavier ones. For instance, if you've ever noticed the scent of ammonia spreading quickly in a room, it's because of its relatively low molar mass.

In the context of Graham's Law of Diffusion, the rate at which a gas diffuses is inversely proportional to the square root of its molar mass. This might sound a bit technical, but we can simplify it. Let's say you have two gases, Gas A and Gas B. If Gas A is lighter than Gas B, Gas A will diffuse faster as per Graham's Law. Mathematically, the relationship can be expressed as:

• \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \)

This formula helps you compare the diffusion rates of two gases when you know their molar masses.

Molar Mass Calculation

Calculating the molar mass of a gas is crucial for applying Graham's Law effectively. The molar mass refers to the mass of one mole of a substance, usually measured in grams per mole (g/mol). When it comes to gases, this number is vital as it directly influences the rate of diffusion.

To calculate the molar mass, you start by identifying the molecular formula of the gas. For example, oxygen is \( O_2 \) and has a molar mass of approximately 32 g/mol because each oxygen atom contributes 16 g/mol to the molecule. Once you know the molecular formula, you add up the atomic masses to find the total molar mass.

• For atomic masses, refer to the periodic table.
• Sum the atomic masses based on the number of each type of atom in the molecule.

This calculation becomes especially handy when you need to figure out how one gas's diffusion rate compares to another's, allowing you to solve numerous practical and theoretical problems in chemistry.

Gas Diffusion Comparison

Comparing the diffusion rates of different gases is not just a textbook exercise—it's a practical application seen in various scientific and industrial processes. Using Graham's Law of Diffusion, we can predict how fast a gas will diffuse relative to another.

To compare two gases:

• Identify the molar masses of both gases.
• Use the formula \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \) to calculate the rate of diffusion.

In our provided problem, the unknown gas diffuses four times as fast as oxygen. Given that oxygen's molar mass is 32 g/mol, you can set up the equation: \( \frac{4r_2}{r_2}= \sqrt{\frac{32}{M_2}} \). By solving, you'll find that the unknown gas has a molar mass of 2 g/mol.

Understanding this concept allows scientists and engineers to manipulate gas behavior in environments like chemical plants and laboratories, making it a vital aspect of chemical engineering and atmospheric studies.