The ratio of the rate of diffusion of a given element to that of helium at the same pressure is 1.4. The molecular weight of the element is

Options

A. 2
B. 4
C. 8
D. 16

Answer

To find the molecular weight of the given element using the information provided, 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:
\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \]
where:
- \(\text{Rate}_1\) is the rate of diffusion of gas 1 (the given element),
- \(\text{Rate}_2\) is the rate of diffusion of gas 2 (helium),
- \(M_1\) is the molar mass of gas 1 (the given element),
- \(M_2\) is the molar mass of gas 2 (helium).

### Step 2: Set Up the Equation
According to the problem, the ratio of the rate of diffusion of the given element to that of helium is 1.4. Therefore, we can write:
\[ \frac{\text{Rate of element}}{\text{Rate of helium}} = 1.4 \]
This can be substituted into Graham's Law:
\[ 1.4 = \sqrt{\frac{M_{\text{He}}}{M_{\text{element}}}} \]

### Step 3: Square Both Sides
To eliminate the square root, we square both sides of the equation:
\[ (1.4)^2 = \frac{M_{\text{He}}}{M_{\text{element}}} \]
Calculating \(1.4^2\):
\[ 1.96 = \frac{M_{\text{He}}}{M_{\text{element}}} \]

### Step 4: Substitute the Molar Mass of Helium
The molar mass of helium (\(M_{\text{He}}\)) is approximately 4 g/mol. Substitute this value into the equation:
\[ 1.96 = \frac{4}{M_{\text{element}}} \]

### Step 5: Rearrange to Solve for Molar Mass of the Element
Now, we rearrange the equation to solve for \(M_{\text{element}}\):
\[ M_{\text{element}} = \frac{4}{1.96} \]

### Step 6: Calculate the Molar Mass
Now, perform the division:
\[ M_{\text{element}} \approx 2.04 \text{ g/mol} \]

### Step 7: Round the Result
Since the problem may require a rounded answer, we can round 2.04 to 2 g/mol.

### Final Answer
The molecular weight of the given element is approximately **2 g/mol**.

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