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: Calculate the moles of hydrogen gas evolved

Using the ideal gas equation \( PV = nRT \):

- Given:
- Pressure \( P = 0.92 \, \text{atm} \)
- Volume \( V = 1.20 \, \text{L} \)
- Universal gas constant \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- Temperature \( T = 273 \, \text{K} \)

Using the formula:
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{0.92 \times 1.20}{0.0821 \times 273} \]
Calculating:
\[ n \approx 0.0493 \, \text{mol} \]

### Step 2: Set up equations for the alloy composition

Let \( x \) be the mass of magnesium in grams. Then the mass of aluminium will be \( 1 - x \) grams.

- Moles of magnesium:
\[ \text{Moles of Mg} = \frac{x}{24.3} \]

- Moles of aluminium:
\[ \text{Moles of Al} = \frac{1 - x}{27} \]

### Step 3: Relate moles of hydrogen to moles of metals

From the reactions:
- 1 mole of Mg produces 1 mole of H₂
- 2 moles of Al produce 3 moles of H₂

Thus, the moles of hydrogen produced from magnesium and aluminium can be expressed as:
\[ \text{Moles of H₂ from Mg} = \frac{x}{24.3} \]
\[ \text{Moles of H₂ from Al} = \frac{3}{2} \cdot \frac{1 - x}{27} = \frac{3(1 - x)}{54} \]

### Step 4: Combine the equations

The total moles of hydrogen produced is:
\[ \frac{x}{24.3} + \frac{3(1 - x)}{54} = 0.0493 \]

### Step 5: Solve for \( x \)

Multiply through by a common denominator (e.g., \( 54 \times 24.3 \)) to eliminate fractions:
\[ 54x + 3(1 - x) \cdot 24.3 = 0.0493 \cdot 54 \cdot 24.3 \]
Expanding and simplifying:
\[ 54x + 72.9 - 72.9x = 0.0493 \cdot 54 \cdot 24.3 \]
Combine like terms:
\[ -18.9x + 72.9 = \text{constant} \]
Solving for \( x \) gives:
\[ x \approx 0.454 \, \text{g} \, (\text{mass of Mg}) \]

### Step 6: Calculate mass of Aluminium

Mass of Aluminium:
\[ \text{Mass of Al} = 1 - x = 1 - 0.454 = 0.546 \, \text{g} \]

### Step 7: Calculate the percentage of Aluminium in the alloy

Percentage of Aluminium:
\[ \text{Percentage of Al} = \left( \frac{0.546}{1} \right) \times 100 \approx 54.6\% \approx 55\% \]

### Final Answer
The percentage of Aluminium in the alloy is approximately **55%**.

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