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 when 4g of oxygen diffuses through a narrow hole, we can apply 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. Mathematically, it can be expressed 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
In this case, we have:
- Gas 1: Hydrogen (H₂) with a molar mass \( M_1 = 2 \, \text{g/mol} \)
- Gas 2: Oxygen (O₂) with a molar mass \( M_2 = 32 \, \text{g/mol} \)

### Step 3: Set Up the Equation
Since we know that 4g of oxygen diffuses, we need to find out how much hydrogen would diffuse under the same conditions. We can express the rates of diffusion in terms of the number of moles:
\[ \frac{n_{H_2}}{n_{O_2}} = \frac{M_{O_2}}{M_{H_2}} \]
where \( n \) represents the number of moles.

### Step 4: Calculate the Number of Moles of Oxygen
First, calculate the number of moles of oxygen that corresponds to 4g:
\[ n_{O_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \, \text{g}}{32 \, \text{g/mol}} = 0.125 \, \text{mol} \]

### Step 5: Use Graham's Law to Find Moles of Hydrogen
Using Graham's Law:
\[ \frac{n_{H_2}}{0.125} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \]
Thus,
\[ n_{H_2} = 4 \times 0.125 = 0.5 \, \text{mol} \]

### Step 6: Calculate the Mass of Hydrogen
Now, convert the number of moles of hydrogen back to grams:
\[ \text{mass of } H_2 = n_{H_2} \times M_{H_2} = 0.5 \, \text{mol} \times 2 \, \text{g/mol} = 1 \, \text{g} \]

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

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