An element undergoes a reaction as shown :
`X+e^(-) to X^(-)` Energy released =30.876 eV
The energy released, is used to dissociate 8 g of `H_2` molecules equality into `H^+` and `H^*`, where `H^*` is in an excited state, in which the electron travels a path length equal to four times its debroglie wavelength.
(a)Determine the least amount (moles) of 'X' that would be required.
Given : I.E. of H =13.6eV/atom
Bond energy of `H_2` =4.526 eV/molecule.

Explanation

To solve the problem step by step, we need to determine the least amount of moles of element 'X' required to dissociate 8 g of hydrogen molecules into \( H^+ \) and \( H^* \).

### Step 1: Calculate the number of moles of \( H_2 \)

Given that the molecular mass of \( H_2 \) is 2 g/mol, we can calculate the number of moles of \( H_2 \) in 8 g.

\[ \text{Number of moles of } H_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \, \text{g}}{2 \, \text{g/mol}} = 4 \, \text{moles} \]

**Hint for Step 1:** Remember that the molar mass of \( H_2 \) is 2 g/mol, which is essential for calculating moles.

### Step 2: Calculate the total energy required to dissociate \( H_2 \)

The total energy required to dissociate 4 moles of \( H_2 \) into \( H^+ \) and \( H^* \) involves two components: the bond energy of \( H_2 \) and the ionization energy of hydrogen.

1. **Bond energy of \( H_2 \)**: 4.526 eV/molecule
2. **Ionization energy of \( H \)**: 13.6 eV/atom

Total energy required for 4 moles of \( H_2 \):

\[ \text{Total energy required} = \text{(Bond energy)} + \text{(Ionization energy)} \]

Calculating the total energy:

\[ \text{Total energy required} = 4 \times N_A \times 4.526 + 4 \times N_A \times 13.6 \]

Where \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \)).

**Hint for Step 2:** Make sure to multiply the energy per molecule by the total number of molecules, which is given by \( \text{moles} \times N_A \).

### Step 3: Calculate the energy released by the reaction

The energy released from the reaction of element \( X \) with an electron is given as 30.876 eV. If \( x \) is the number of moles of \( X \), the total energy released can be expressed as:

\[ \text{Total energy released} = x \times N_A \times 30.876 \]

**Hint for Step 3:** Remember that the energy released is dependent on the number of moles of \( X \) and the energy per reaction.

### Step 4: Set up the equation

Since the total energy released must equal the total energy required, we can set up the equation:

\[ x \times N_A \times 30.876 = 4 \times N_A \times 4.526 + 4 \times N_A \times 13.6 \]

**Hint for Step 4:** Notice that \( N_A \) cancels out from both sides of the equation, simplifying the calculation.

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

Cancelling \( N_A \) and rearranging gives:

\[ x \times 30.876 = 4 \times 4.526 + 4 \times 13.6 \]

Calculating the right side:

\[ 4 \times 4.526 = 18.104 \quad \text{and} \quad 4 \times 13.6 = 54.4 \]
\[ \text{Total energy required} = 18.104 + 54.4 = 72.504 \]

Now, substituting back into the equation:

\[ x \times 30.876 = 72.504 \]

Solving for \( x \):

\[ x = \frac{72.504}{30.876} \approx 2.35 \]

### Step 6: Round to the least whole number of moles

Since we need the least amount of moles of \( X \), we round up to the nearest whole number:

\[ \text{Least amount of moles of } X = 3 \]

**Final Answer:** The least amount of moles of \( X \) required is approximately 3.