A mixture of `O_(2)` and gas `"Y"` `( mol. wt. 80)` in the mole ratio `a:b` has a mean molecular weight 40. What would be mean molecular weight, if the gases are mixed in the ratio `b:a` under identical conditions ? ( gases are )

Options

A. 40
B. 48
C. 62
D. 72

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the given data
We have a mixture of \( O_2 \) (oxygen) and a gas \( Y \) with a molecular weight of 80. The mole ratio of \( O_2 \) to \( Y \) is given as \( a:b \) and the mean molecular weight of the mixture is 40.

### Step 2: Set up the equation for mean molecular weight
The mean molecular weight \( M \) of a gas mixture can be calculated using the formula:
\[ M = (M_1 \cdot X_1) + (M_2 \cdot X_2) \]
where \( M_1 \) and \( M_2 \) are the molecular weights of the gases and \( X_1 \) and \( X_2 \) are their respective mole fractions.

For our case:
- \( M_1 = 32 \) g/mol (molecular weight of \( O_2 \))
- \( M_2 = 80 \) g/mol (molecular weight of gas \( Y \))
- Mean molecular weight \( M = 40 \)

Let \( X_1 \) be the mole fraction of \( O_2 \) and \( X_2 \) be the mole fraction of gas \( Y \). Since \( X_1 + X_2 = 1 \), we can express \( X_2 \) as \( 1 - X_1 \).

### Step 3: Substitute into the mean molecular weight equation
Substituting the known values into the equation:
\[ 40 = (32 \cdot X_1) + (80 \cdot (1 - X_1)) \]

### Step 4: Solve for \( X_1 \)
Expanding the equation:
\[ 40 = 32X_1 + 80 - 80X_1 \]
\[ 40 = 80 - 48X_1 \]
Rearranging gives:
\[ 48X_1 = 80 - 40 \]
\[ 48X_1 = 40 \]
\[ X_1 = \frac{40}{48} = \frac{5}{6} \]

### Step 5: Determine the mole fraction of gas \( Y \)
Using \( X_2 = 1 - X_1 \):
\[ X_2 = 1 - \frac{5}{6} = \frac{1}{6} \]

### Step 6: Find the new mean molecular weight when the ratio is changed
Now, if the gases are mixed in the ratio \( b:a \) (which means \( Y \) to \( O_2 \)), the new mole fractions will be:
- \( X_1' = \frac{1}{6} \) (for \( O_2 \))
- \( X_2' = \frac{5}{6} \) (for gas \( Y \))

### Step 7: Calculate the new mean molecular weight
Using the new mole fractions in the mean molecular weight equation:
\[ M' = (32 \cdot X_1') + (80 \cdot X_2') \]
Substituting the values:
\[ M' = (32 \cdot \frac{1}{6}) + (80 \cdot \frac{5}{6}) \]
Calculating each term:
\[ M' = \frac{32}{6} + \frac{400}{6} = \frac{432}{6} = 72 \]

### Final Answer
The mean molecular weight when the gases are mixed in the ratio \( b:a \) is **72 g/mol**.

---