Question

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

Short answer

The molar weight of the gas is 2 g/mol (option a).

Step-by-step solution

Understanding the Problem

The problem states that a certain gas diffuses four times as fast as oxygen. We need to determine the molar weight of this gas. We'll use Graham's law of effusion to solve this.

Applying Graham's Law of Effusion

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. It can be described by the formula: \( \frac{\text{Rate of Gas 1}}{\text{Rate of Gas 2}} = \sqrt{\frac{M_2}{M_1}} \), where \( M_1 \) and \( M_2 \) are the molar masses of gas 1 and gas 2 respectively.

Setting Up the Equation

Since the unknown gas diffuses four times as quickly as oxygen, we set the ratio of their rates as 4:1 and oxygen as the second gas. Therefore, \( \frac{4}{1} = \sqrt{\frac{32}{M}} \), where \( M \) is the molar mass of the unknown gas and the molar mass of oxygen is 32 g/mol.

Solving the Equation

First, square both sides of the equation to eliminate the square root: \( 16 = \frac{32}{M} \). Then, solve for \( M \) by multiplying both sides by \( M \): \( M \times 16 = 32 \). Thus, \( M = \frac{32}{16} \).

Finding the Molar Mass

Simplify \( \frac{32}{16} \) to find that \( M = 2 \). Therefore, the molar mass of the gas is 2 g/mol.

Key concepts

Rate of Diffusion

Understanding the rate of diffusion is key to determining how quickly a gas spreads compared to another. In this exercise, we deal with a gas that diffuses four times faster than oxygen. The rate of diffusion tells us how quickly molecules spread from an area of high concentration to one of low concentration. The factors influencing diffusion include:

• Temperature: Higher temperatures increase kinetic energy and speed up diffusion.
• Molecular Size: Smaller molecules diffuse faster than larger ones.
• Concentration Gradient: A steeper gradient increases the diffusion rate.

Graham's Law states that the diffusion rate of a gas is inversely proportional to the square root of its molar mass. Therefore, lighter gases diffuse more quickly because their molecules move faster, leading to more collisions and quicker spread.

Molar Mass Calculation

Calculating molar mass helps us determine the mass of one mole of a substance. In this problem, finding the molar mass of an unknown gas is essential to understanding its diffusion characteristics. Here, we are given:

• Oxygen's molar mass: 32 g/mol
• The unknown gas diffuses four times faster than oxygen.

Using the formula derived from Graham's Law: \[ \frac{4}{1} = \sqrt{\frac{32}{M}} \]We square both sides to remove the square root: \[ 16 = \frac{32}{M} \]Solving this equation helps us isolate the molar mass \( M \):Multiply both sides by \( M \) to get:\[ 16M = 32 \]Finally, dividing both sides by 16, we find:\[ M = 2 \]This calculation reveals that the molar mass of the unknown gas is 2 g/mol, indicating it is lighter and thus diffuses faster than oxygen.

Effusion Rate Comparison

Effusion involves the escape of gas molecules through a tiny opening into a vacuum. The rate at which this occurs also follows Graham's Law. Here, we compare the effusion rates of the unknown gas to oxygen.Graham's Law of effusion shows that:

• The effusion rate is directly related to the speed of molecules.
• Lighter gases, like the unknown gas with a molar mass of 2 g/mol, effuse faster compared to heavier gases like oxygen.

The law provides the relationship:\[ \frac{\text{Rate of Gas 1}}{\text{Rate of Gas 2}} = \sqrt{\frac{M_2}{M_1}} \]Using the known diffusion rate (4 times faster than oxygen) and oxygen's molar mass (32 g/mol), we can analyze:The unknown gas has a molar mass of 2 g/mol, allowing it to effuse faster as its molecules are smaller and have higher velocities.Thus, understanding the concept of effusion rates illuminates the underlying reasons why gases with lower molar mass move more quickly, reinforcing the solution's correctness of the molar mass calculation.