If 4g of oxygen diffuses through a very narrow hole, how much hydrogen would have diffused under identical conditions?

Options

A. 16 g
B. 1 g
C. `1//4` g
D. 64 g

Explanation

To solve the problem of how much hydrogen would diffuse if 4 grams of oxygen diffuses through a narrow hole under identical conditions, we can use Graham's law of diffusion. Here’s a 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. This can be expressed mathematically as:
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( r_1 \) and \( r_2 \) are the rates of diffusion of gases 1 and 2, and \( M_1 \) and \( M_2 \) are their molar masses.

### Step 2: Identify the Gases and Their Molar Masses
In this case:
- Gas 1 (Oxygen, O₂): Molar mass \( M_1 = 32 \, \text{g/mol} \)
- Gas 2 (Hydrogen, H₂): Molar mass \( M_2 = 2 \, \text{g/mol} \)

### Step 3: Set Up the Rate of Diffusion Equation
Let \( x \) be the mass of hydrogen that diffuses. According to Graham's law:
\[ \frac{r_{\text{H}_2}}{r_{\text{O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{H}_2}}} \]
This can be rewritten in terms of the masses:
\[ \frac{\frac{x}{2}}{\frac{4}{32}} = \sqrt{\frac{32}{2}} \]

### Step 4: Simplify the Equation
The left side simplifies to:
\[ \frac{x}{2} \div \frac{4}{32} = \frac{x}{2} \times \frac{32}{4} = \frac{8x}{2} = 4x \]
The right side simplifies to:
\[ \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \]
Thus, we have:
\[ 4x = 4 \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{4}{4} = 1 \, \text{g} \]

### Conclusion
Therefore, the mass of hydrogen that would have diffused under identical conditions is **1 gram**.

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