A gaseous mixture containing equal mole sof `H_(2),O_(2)` and He is subjected to series of effusion steps. The composition (by moles) of effused mixture after 4 effusion steps is `x:1:y` rspectively. Then find the value of `((x)/(y)).`

Explanation

To solve the problem, we need to analyze the effusion of the gaseous mixture containing equal moles of \( H_2 \), \( O_2 \), and \( He \). The key concept we will use is Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify the Molar Masses**:
- Molar mass of \( H_2 \) = 2 g/mol
- Molar mass of \( O_2 \) = 32 g/mol
- Molar mass of \( He \) = 4 g/mol

2. **Calculate the Rate of Effusion**:
According to Graham's law, the rate of effusion (r) is given by:
\[ r \propto \frac{1}{\sqrt{M}} \]
where \( M \) is the molar mass of the gas.

- For \( H_2 \):
\[ r_{H_2} \propto \frac{1}{\sqrt{2}} \]
- For \( O_2 \):
\[ r_{O_2} \propto \frac{1}{\sqrt{32}} = \frac{1}{4\sqrt{2}} \]
- For \( He \):
\[ r_{He} \propto \frac{1}{\sqrt{4}} = \frac{1}{2} \]

3. **Determine the Ratio of Effusion Rates**:
The ratio of the rates of effusion of \( H_2 \), \( O_2 \), and \( He \) can be expressed as:
\[ r_{H_2} : r_{O_2} : r_{He} = \frac{1}{\sqrt{2}} : \frac{1}{4\sqrt{2}} : \frac{1}{2} \]

To simplify this, we can express each term in a common format:
\[ = \frac{1}{\sqrt{2}} : \frac{1}{4\sqrt{2}} : \frac{1}{2} = 4 : 1 : 2 \]

4. **Calculate the Composition after 4 Effusion Steps**:
After \( n \) effusion steps, the composition of the gases will change according to the ratio of their effusion rates. After 4 effusion steps, the composition will be:
\[ x : 1 : y = (4)^{4} : (1)^{4} : (2)^{4} = 256 : 1 : 16 \]

5. **Find the Value of \( \frac{x}{y} \)**:
From the composition \( x : 1 : y = 256 : 1 : 16 \):
\[ x = 256, \quad y = 16 \]
Therefore,
\[ \frac{x}{y} = \frac{256}{16} = 16 \]

### Final Answer:
The value of \( \frac{x}{y} \) is \( 16 \).

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