Three different gases X, Y and Z of molecular masses 2, 16 and 64 were enclosed in a vessel at constant temperature till equilbrium is reaches. Which of the following statement is correct?

Options

A. Gas Z will be at the top of the vessel
B. Gas Y will be at the top of the vessel
C. Gas X will be at the bottom and Z will be at the top
D. Gases wil form homogeneous mixture

Answer

To solve the question regarding the behavior of three different gases (X, Y, and Z) with molecular masses 2, 16, and 64, we can follow these steps:

### Step 1: Understand the Concept of Gas Mixtures
Gases in a mixture behave independently of each other. This means that when different gases are mixed, they will not form layers based on their molecular masses. Instead, they will distribute evenly throughout the container.

### Step 2: Recognize the Homogeneous Nature of Gas Mixtures
In a gas mixture, the gases will spread out to fill the entire volume of the container uniformly. This is due to the random motion of gas molecules, which leads to a homogeneous mixture regardless of their individual molecular masses.

### Step 3: Analyze the Given Gases
- Gas X has a molecular mass of 2 (likely hydrogen).
- Gas Y has a molecular mass of 16 (likely oxygen).
- Gas Z has a molecular mass of 64 (likely sulfur hexafluoride).

Despite their different molecular masses, all three gases will mix uniformly in the vessel.

### Step 4: Evaluate the Statements
Given the understanding of gas behavior, we can conclude that:
- No gas will be at the top or bottom of the vessel based on its molecular mass.
- All gases will be evenly distributed throughout the vessel.

### Conclusion
The correct statement is that the mixture of gases (X, Y, and Z) will be a uniform homogeneous mixture.

### Final Answer
The correct statement is: "The gases form a uniform homogeneous mixture."

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