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

Explanation

To solve the problem, we need to find the mean molecular weight of a mixture of gases when the ratio of the gases is changed. Let's break down the solution step by step.

### Step 1: Define the known values
- Molar mass of \( O_2 \) = 32 g/mol
- Molar mass of gas \( Y \) = 80 g/mol
- Mean molecular weight of the mixture when the ratio is \( a:b \) = 40 g/mol

### Step 2: Set up the equation for the mean molecular weight
The mean molecular weight \( M \) of a mixture can be calculated using the formula:
\[ M = \frac{(M_1 \cdot n_1) + (M_2 \cdot n_2)}{n_1 + n_2} \]
Where:
- \( M_1 \) = molar mass of \( O_2 \) = 32 g/mol
- \( M_2 \) = molar mass of gas \( Y \) = 80 g/mol
- \( n_1 \) = number of moles of \( O_2 \) = \( a \)
- \( n_2 \) = number of moles of gas \( Y \) = \( b \)

Thus, we can write:
\[ 40 = \frac{(32a + 80b)}{a + b} \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 40(a + b) = 32a + 80b \]
Expanding this:
\[ 40a + 40b = 32a + 80b \]

### Step 4: Rearranging the equation
Rearranging the terms to isolate \( a \) and \( b \):
\[ 40a - 32a = 80b - 40b \]
This simplifies to:
\[ 8a = 40b \]
Dividing both sides by 8:
\[ a = 5b \]

### Step 5: Determine the ratio of \( a \) to \( b \)
From \( a = 5b \), we can express the ratio \( \frac{a}{b} = 5 \). Therefore, the ratio \( a:b = 5:1 \).

### Step 6: Find the mean molecular weight for the new ratio \( b:a \)
Now, we need to find the mean molecular weight when the gases are mixed in the ratio \( b:a \) (which is \( 1:5 \)):
- \( n_1 = b = 1 \)
- \( n_2 = a = 5 \)

Using the mean molecular weight formula again:
\[ M' = \frac{(M_1 \cdot n_1) + (M_2 \cdot n_2)}{n_1 + n_2} \]
Substituting the values:
\[ M' = \frac{(32 \cdot 1) + (80 \cdot 5)}{1 + 5} \]
Calculating the numerator:
\[ M' = \frac{32 + 400}{6} = \frac{432}{6} = 72 \]

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

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