Problem 143 - A Complete Resource Book in Chemistry for JEE Main

Question

\(4 g m\) of oxygen diffuses through a very narrow hole in \(30 \mathrm{sec}\). What mass of hydrogen in gm will diffuse in the same time under identical conditions?

Step-by-Step Solution

Answer

16 g of hydrogen will diffuse in 30 seconds under identical conditions.

Step 1: Understand Graham's Law of Diffusion

Graham's law of diffusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, \( \text{Rate of Effusion} \propto \frac{1}{\sqrt{M}} \) where \( M \) is the molar mass of the gas.

Step 2: Set up the Ratio Using Graham's Law

According to Graham's law, for two gases A and B, \( \frac{R_A}{R_B} = \sqrt{\frac{M_B}{M_A}} \), where \( R_A \) and \( R_B \) are the rates (amount per unit time) of effusion, and \( M_A \) and \( M_B \) are their molar masses. Here, \( M_A = 32 \ g/mol \) for oxygen, and \( M_B = 2 \ g/mol \) for hydrogen.

Step 3: Calculate the Rate of Diffusion for Oxygen

The rate of diffusion for oxygen, \( R_O \), is \( \frac{4 \ g}{30 \ s} = \frac{2}{15} \ g/s \).

Step 4: Solve for the Rate of Hydrogen Diffusion

Using the ratio \( \frac{R_H}{R_O} = \sqrt{\frac{M_O}{M_H}} \), where \(R_H\) is the rate of hydrogen: \( R_H = R_O \cdot \sqrt{\frac{M_O}{M_H}} = \frac{2}{15} \cdot \sqrt{\frac{32}{2}} \).

Step 5: Compute Hydrogen Diffusion Rate

Simplify \( R_H = \frac{2}{15} \cdot \sqrt{16} = \frac{2}{15} \cdot 4 = \frac{8}{15} \ g/s \).

Step 6: Calculate Total Mass of Hydrogen Diffused

The total mass \( m_H \) diffused over \( 30 \ s \) is \( R_H \times 30 = \frac{8}{15} \times 30 = 16 \ g \).

Key Concepts

Rate of Effusion

The rate of effusion refers to how fast a gas escapes through a small hole into a vacuum. According to Graham's Law, this rate is inversely related to the square root of the gas's molar mass. To put it simply, lighter gases effuse faster than heavier ones. By understanding this principle, we can predict how quickly different gases will pass through a tiny opening.
For instance:

• Lighter gases like hydrogen will move rapidly, resulting in a high rate of effusion.
• Heavier gases like oxygen will move more slowly, resulting in a lower rate of effusion.

In the exercise, we see the effusion of oxygen calculated as a rate of grams per second, showcasing this principle in action.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It plays a crucial role in determining the rate of effusion of gases. According to Graham's Law, the smaller the molar mass of a gas, the higher its rate of effusion.

• For oxygen, the molar mass is 32 g/mol, which is considered relatively high compared to lighter gases.
• Hydrogen, with a molar mass of 2 g/mol, is much lighter and will effuse much more quickly.

In our exercise, by comparing the molar masses of oxygen and hydrogen, we used Graham's Law to predict how much faster hydrogen would diffuse than oxygen.

Oxygen and Hydrogen Diffusion

Diffusion describes how molecules spread from areas of high concentration to areas of low concentration. This natural process can be dramatically different for different gases based on their properties.
Oxygen and hydrogen, with vastly different molar masses, demonstrate a marked contrast in their diffusion rates through narrow openings.

• Even though both gases are under identical conditions, hydrogen will diffuse much faster because of its lower molar mass.
• Hydrogen's rate of diffusion is calculated to be four times faster than oxygen, highlighting this stark difference.

By understanding these differences, students can gain insight into practical applications, such as separating gases or predicting how quickly gases will escape from containers.