The ionization enthalpies of Li and Na are `"520 kJ mol"^(-1)` and `"495 kJ mol"^(-1)` respectively. They energy required to convert all the atoms present in 70 mg of Li vapours and 230 mg of sodium vapours of their respectively gaseous cations respectively are

Options

A. 5200 J, 4950 J
B. 4950 J, 5200 J
C. 52 J, 49.5 J
D. 495 J, 520 J

Answer

To solve the problem, we need to calculate the energy required to ionize the given amounts of lithium (Li) and sodium (Na) vapors. We will use the ionization enthalpies provided and the molar masses of the elements to find the solution step by step.

### Step 1: Calculate the number of moles of Lithium (Li)

Given:
- Mass of Li = 70 mg = 0.070 g
- Molar mass of Li = 7 g/mol

To find the number of moles of Li, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

\[ \text{Number of moles of Li} = \frac{0.070 \text{ g}}{7 \text{ g/mol}} = 0.010 \text{ moles} \]

### Step 2: Calculate the energy required to ionize Lithium (Li)

The ionization enthalpy of Li is given as 520 kJ/mol. Therefore, the energy required for 0.010 moles of Li is:

\[ \text{Energy for Li} = \text{Number of moles} \times \text{Ionization enthalpy} \]

\[ \text{Energy for Li} = 0.010 \text{ moles} \times 520 \text{ kJ/mol} = 5.2 \text{ kJ} \]

### Step 3: Calculate the number of moles of Sodium (Na)

Given:
- Mass of Na = 230 mg = 0.230 g
- Molar mass of Na = 23 g/mol

Using the same formula for the number of moles:

\[ \text{Number of moles of Na} = \frac{0.230 \text{ g}}{23 \text{ g/mol}} = 0.010 \text{ moles} \]

### Step 4: Calculate the energy required to ionize Sodium (Na)

The ionization enthalpy of Na is given as 495 kJ/mol. Therefore, the energy required for 0.010 moles of Na is:

\[ \text{Energy for Na} = \text{Number of moles} \times \text{Ionization enthalpy} \]

\[ \text{Energy for Na} = 0.010 \text{ moles} \times 495 \text{ kJ/mol} = 4.95 \text{ kJ} \]

### Step 5: Calculate the total energy required

Now, we add the energies required for both Li and Na:

\[ \text{Total Energy} = \text{Energy for Li} + \text{Energy for Na} \]

\[ \text{Total Energy} = 5.2 \text{ kJ} + 4.95 \text{ kJ} = 10.15 \text{ kJ} \]

### Step 6: Convert the total energy to Joules

Since 1 kJ = 1000 J, we convert the total energy:

\[ \text{Total Energy in Joules} = 10.15 \text{ kJ} \times 1000 = 10150 \text{ J} \]

### Final Answer

The energy required to convert all the atoms present in 70 mg of Li vapors and 230 mg of sodium vapors into their respective gaseous cations is **10150 J**.

---