The first second and third ionisation energies of AI are 587 , 1817 and 2745 kJ `"mol"^(-1)` respectively . Calculate the energy required to convert all the atoms of AI to `AI^(+3)` present in 270 mg of AI vapours

Options

A. 5149kJ
B. 51.49 kJ
C. 2745 kJ
D. 514.9 kJ

Answer

To solve the problem of calculating the energy required to convert all the atoms of aluminum (Al) to Al³⁺ in 270 mg of Al vapors, we will follow these steps:

### Step 1: Understand Ionization Energies
The first, second, and third ionization energies of aluminum are given as:
- First ionization energy (IE₁) = 587 kJ/mol
- Second ionization energy (IE₂) = 1817 kJ/mol
- Third ionization energy (IE₃) = 2745 kJ/mol

### Step 2: Calculate Total Ionization Energy
To convert one mole of Al to Al³⁺, we need to sum the three ionization energies:
\[ \text{Total Ionization Energy} = IE₁ + IE₂ + IE₃ \]
Substituting the values:
\[ \text{Total Ionization Energy} = 587 + 1817 + 2745 = 5149 \text{ kJ/mol} \]

### Step 3: Determine the Molar Mass of Aluminum
The molar mass of aluminum (Al) is approximately 27 g/mol.

### Step 4: Convert Mass of Aluminum to Moles
We need to convert 270 mg of aluminum to grams:
\[ 270 \text{ mg} = 0.270 \text{ g} \]
Now, convert grams to moles using the molar mass:
\[ \text{Moles of Al} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.270 \text{ g}}{27 \text{ g/mol}} = 0.010 \text{ mol} \]

### Step 5: Calculate Energy Required for 0.010 moles
The energy required to convert 1 mole of Al to Al³⁺ is 5149 kJ. Therefore, for 0.010 moles:
\[ \text{Energy required} = \text{Moles} \times \text{Total Ionization Energy} = 0.010 \text{ mol} \times 5149 \text{ kJ/mol} = 51.49 \text{ kJ} \]

### Final Answer
The energy required to convert all the atoms of aluminum in 270 mg to Al³⁺ is **51.49 kJ**.

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