A mixture of methane and ethane in the molar ratio of x:y has a mean molar mass of 20. what would be the mean molar mass, if the gases are mixed in the molar ratio of y:x?

Options

A. 20
B. 26
C. 24
D. 15

Answer

To solve the problem step by step, we will analyze the given information and apply the necessary calculations.

### Step 1: Understand the given data
We have a mixture of methane (CH₄) and ethane (C₂H₆) in a molar ratio of \(x:y\) with a mean molar mass of 20 g/mol.

### Step 2: Write the molar masses of the components
- Molar mass of methane (CH₄) = 16 g/mol
- Molar mass of ethane (C₂H₆) = 30 g/mol

### Step 3: Set up the equation for mean molar mass
The mean molar mass (M) of the mixture can be expressed as:
\[ M = \frac{(16x + 30y)}{(x + y)} = 20 \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ 16x + 30y = 20(x + y) \]

### Step 5: Expand and rearrange the equation
Expanding the right side:
\[ 16x + 30y = 20x + 20y \]
Rearranging gives:
\[ 30y - 20y = 20x - 16x \]
This simplifies to:
\[ 10y = 4x \]

### Step 6: Solve for the ratio of x to y
From \(10y = 4x\), we can express \(x\) in terms of \(y\):
\[ x = \frac{10}{4}y = \frac{5}{2}y \]

### Step 7: Find the mean molar mass for the new ratio \(y:x\)
Now, we need to find the mean molar mass when the gases are mixed in the molar ratio \(y:x\) (which is \(y:\frac{5}{2}y\) or \(2:5\)).

### Step 8: Set up the new mean molar mass equation
Let’s denote the new mean molar mass as \(M'\):
\[ M' = \frac{(16y + 30(\frac{5}{2}y))}{(y + \frac{5}{2}y)} \]

### Step 9: Simplify the equation
Calculating the numerator:
\[ M' = \frac{(16y + 75y)}{(y + 2.5y)} = \frac{91y}{3.5y} \]

### Step 10: Cancel out \(y\) and simplify
\[ M' = \frac{91}{3.5} = 26 \text{ g/mol} \]

### Final Answer
Thus, the mean molar mass when the gases are mixed in the molar ratio \(y:x\) is **26 g/mol**.

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