1.0 g of an alloy of Al and Mg when treated with excess of dil HCl gave `MgCl_(2), AlCl_(3)` and hydrogen. Evolved hydrogen collected over Hg at `0^(@)C` has a volume of 1.20 litres at 0.92 atm pressure. The percentage of Aluminium in the alloy is :

Options

A. `15%`
B. `55%`
C. `46%`
D. `25%`

Explanation

To find the percentage of Aluminium in the alloy, we will follow these steps:

### Step 1: Identify the variables
Let \( x \) be the mass of Magnesium (Mg) in the alloy. Since the total mass of the alloy is 1.0 g, the mass of Aluminium (Al) will be \( 1.0 - x \) g.

### Step 2: Calculate moles of Magnesium and Aluminium
The moles of Magnesium can be calculated using its molar mass:
\[ \text{Moles of Mg} = \frac{x}{24} \]
The moles of Aluminium can be calculated similarly:
\[ \text{Moles of Al} = \frac{1 - x}{27} \]

### Step 3: Write the reactions
The reactions of Magnesium and Aluminium with dilute HCl are:
1. For Magnesium:
\[ \text{Mg} + 2 \text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \]
From this reaction, 1 mole of Mg produces 1 mole of H₂.

2. For Aluminium:
\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \]
From this reaction, 2 moles of Al produce 3 moles of H₂, or 1 mole of Al produces \( \frac{3}{2} \) moles of H₂.

### Step 4: Calculate total moles of hydrogen produced
The total moles of hydrogen produced from both reactions can be expressed as:
\[ \text{Total moles of H}_2 = \frac{x}{24} + \frac{3}{2} \cdot \frac{1 - x}{27} \]
Simplifying the second term:
\[ \text{Total moles of H}_2 = \frac{x}{24} + \frac{3(1 - x)}{54} \]

### Step 5: Use the ideal gas law to find moles of hydrogen
Using the ideal gas law \( PV = nRT \):
\[ n = \frac{PV}{RT} \]
Given:
- \( P = 0.92 \, \text{atm} \)
- \( V = 1.20 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 273 \, \text{K} \)

Calculating the moles of hydrogen:
\[ n = \frac{0.92 \times 1.20}{0.0821 \times 273} \approx 0.0492 \, \text{mol} \]

### Step 6: Set up the equation
Now we can set up the equation:
\[ \frac{x}{24} + \frac{3(1 - x)}{54} = 0.0492 \]

### Step 7: Solve for \( x \)
Multiply through by the least common multiple (LCM) of the denominators (which is 108):
\[ \frac{108x}{24} + \frac{3 \cdot 108(1 - x)}{54} = 0.0492 \cdot 108 \]
This simplifies to:
\[ 4.5x + 6(1 - x) = 5.3136 \]
\[ 4.5x + 6 - 6x = 5.3136 \]
\[ -1.5x + 6 = 5.3136 \]
\[ -1.5x = -0.6864 \]
\[ x = \frac{0.6864}{1.5} \approx 0.4576 \, \text{g} \]

### Step 8: Calculate mass of Aluminium
The mass of Aluminium in the alloy is:
\[ 1.0 - x = 1.0 - 0.4576 \approx 0.5424 \, \text{g} \]

### Step 9: Calculate percentage of Aluminium
The percentage of Aluminium in the alloy is:
\[ \text{Percentage of Al} = \left( \frac{0.5424}{1.0} \right) \times 100 \approx 54.24\% \]

### Final Answer
The percentage of Aluminium in the alloy is approximately **54%**.

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