Rate of effusion of LPG (a mixture of `n`-butane and propane) is 1.25 times that of `SO_(3)`. Hence, mole fraction of `n`-butane in LPG is

Options

A. 0.75
B. 0.25
C. 0.5
D. 0.67

Answer

To solve the problem of finding the mole fraction of n-butane in LPG (a mixture of n-butane and propane) given that the rate of effusion of LPG is 1.25 times that of SO3, we can follow these steps:

### Step 1: Understand the relationship between effusion rates and molar mass
According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This can be expressed mathematically as:
\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \]
where \(M_1\) and \(M_2\) are the molar masses of the two gases.

### Step 2: Set up the equation using the given information
We know that:
\[ \frac{\text{Rate of LPG}}{\text{Rate of SO}_3} = 1.25 \]
Thus, we can write:
\[ 1.25 = \sqrt{\frac{M_{SO_3}}{M_{LPG}}} \]

### Step 3: Calculate the molar mass of SO3
The molar mass of SO3 can be calculated as follows:
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol, and there are 3 oxygen atoms in SO3.

So,
\[ M_{SO_3} = 32 + (16 \times 3) = 32 + 48 = 80 \text{ g/mol} \]

### Step 4: Substitute the molar mass of SO3 into the equation
Now substituting \(M_{SO_3}\) into the equation:
\[ 1.25 = \sqrt{\frac{80}{M_{LPG}}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 1.25^2 = \frac{80}{M_{LPG}} \]
\[ 1.5625 = \frac{80}{M_{LPG}} \]

### Step 6: Rearrange to find the molar mass of LPG
Rearranging the equation to solve for \(M_{LPG}\):
\[ M_{LPG} = \frac{80}{1.5625} = 51.2 \text{ g/mol} \]

### Step 7: Set up the equation for the mole fraction
Let \(x\) be the mole fraction of n-butane (C4H10), and \(1-x\) be the mole fraction of propane (C3H8). The molar masses are:
- Molar mass of n-butane (C4H10) = \(4 \times 12 + 10 = 58 \text{ g/mol}\)
- Molar mass of propane (C3H8) = \(3 \times 12 + 8 = 44 \text{ g/mol}\)

We can express the average molar mass of the mixture as:
\[ M_{LPG} = x \cdot M_{C4H10} + (1-x) \cdot M_{C3H8} \]
Substituting the known values:
\[ 51.2 = x \cdot 58 + (1-x) \cdot 44 \]

### Step 8: Solve for x
Expanding the equation:
\[ 51.2 = 58x + 44 - 44x \]
\[ 51.2 = 14x + 44 \]
Subtracting 44 from both sides:
\[ 7.2 = 14x \]
Dividing both sides by 14:
\[ x = \frac{7.2}{14} = 0.5142857 \approx 0.5 \]

### Step 9: Conclusion
Thus, the mole fraction of n-butane in LPG is approximately \(0.5\).