The electron affinity of a hypothetical element `'A'` is `3eV` per atom. How much energy in kcal is released when `10g` of `'A'` is completely converted of `A^(-)` ion in a gaseous state ?
`(1 eV=23" kcal mol"^(-1)`, Molar mass of `A=30g)`

Options

A. `23 kcal`
B. `46 kcal`
C. `50kcal`
D. `52kcal`

Explanation

To solve the problem, we need to calculate the energy released when 10 grams of the hypothetical element 'A' is converted to its anion \( A^{-} \) in a gaseous state, given that the electron affinity of 'A' is 3 eV per atom. We will also convert this energy into kilocalories.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Electron affinity of 'A' = 3 eV/atom
- Molar mass of 'A' = 30 g/mol
- Amount of 'A' = 10 g
- Conversion factor: \( 1 \text{ eV} = 23 \text{ kcal/mol} \)

2. **Calculate the Number of Moles of 'A':**
\[ \text{Number of moles of } A = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{30 \text{ g/mol}} = \frac{1}{3} \text{ mol} \]

3. **Calculate the Total Energy Released in eV:**
Since the electron affinity is given per atom, we need to find the total energy for all the moles of 'A':
\[ \text{Total energy in eV} = \text{Number of moles} \times \text{Avogadro's number} \times \text{Electron affinity} \]
\[ \text{Total energy in eV} = \left(\frac{1}{3} \text{ mol}\right) \times (6.022 \times 10^{23} \text{ atoms/mol}) \times 3 \text{ eV/atom} \]
\[ = \frac{1}{3} \times 6.022 \times 10^{23} \times 3 \]
\[ = 6.022 \times 10^{23} \text{ eV} \text{ (since the factor of } \frac{1}{3} \text{ cancels with the 3)} \]

4. **Convert Energy from eV to kcal:**
Now we convert the total energy from eV to kcal using the conversion factor:
\[ \text{Total energy in kcal} = \text{Total energy in eV} \times \frac{23 \text{ kcal}}{1 \text{ eV}} \]
\[ = 6.022 \times 10^{23} \times 23 \text{ kcal} \]
However, since we are interested in the energy per mole, we will consider the energy for the moles calculated earlier:
\[ \text{Total energy in kcal} = \left(\frac{1}{3} \text{ mol}\right) \times 3 \text{ eV/atom} \times 23 \text{ kcal/mol} \]
\[ = 1 \times 23 \text{ kcal} = 23 \text{ kcal} \]

5. **Final Answer:**
The energy released when 10 grams of 'A' is completely converted to \( A^{-} \) ions in a gaseous state is **23 kcal**.