Two grams of dihydrogen diffuse from a container in 10 minutes. How many gram of dioxygen would diffuse through the same container in the same time under similar conditions?

Options

A. 0.5g
B. 4g
C. 6g
D. 8g

Answer

To solve the problem of how many grams of dioxygen would diffuse through the same container in the same time under similar conditions, we can use Graham's law of diffusion. Here’s the step-by-step solution:

### 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. Mathematically, this can be expressed as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where:
- \( R_1 \) = rate of diffusion of dihydrogen (H₂)
- \( R_2 \) = rate of diffusion of dioxygen (O₂)
- \( M_1 \) = molar mass of dihydrogen
- \( M_2 \) = molar mass of dioxygen

### Step 2: Identify Molar Masses
The molar mass of dihydrogen (H₂) is 2 g/mol, and the molar mass of dioxygen (O₂) is 32 g/mol.

### Step 3: Express the Rates of Diffusion
The rate of diffusion can also be expressed in terms of the number of moles of gas diffused over time:
\[ R = \frac{\text{Number of moles}}{\text{Time}} \]
For dihydrogen, we have:
\[ R_1 = \frac{n_1}{t} \]
For dioxygen, we have:
\[ R_2 = \frac{n_2}{t} \]
where \( n_1 \) and \( n_2 \) are the number of moles of H₂ and O₂, respectively, and \( t \) is the time taken for diffusion.

### Step 4: Calculate the Number of Moles of Dihydrogen
Given that 2 grams of dihydrogen diffuse in 10 minutes, we can calculate the number of moles of dihydrogen:
\[ n_1 = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \text{ g}}{2 \text{ g/mol}} = 1 \text{ mol} \]

### Step 5: Set Up the Equation Using Graham's Law
Now substituting \( n_1 \) into Graham's law:
\[ \frac{1/t}{n_2/t} = \sqrt{\frac{32}{2}} \]
This simplifies to:
\[ \frac{1}{n_2} = \sqrt{16} \]
Thus:
\[ n_2 = \frac{1}{4} \text{ mol} \]

### Step 6: Calculate the Mass of Dioxygen
Now, we can find the mass of dioxygen that would diffuse:
\[ \text{mass of O}_2 = n_2 \times M_2 = \frac{1}{4} \text{ mol} \times 32 \text{ g/mol} = 8 \text{ g} \]

### Final Answer
Therefore, 8 grams of dioxygen would diffuse through the same container in the same time under similar conditions.