The ionization energies of Li and Na are `520 kJ mol ^-1 and 495 kJ mol ^-1` respectively. The energy required to convert all the atoms present in 7 mg of Li vapours and 23 mg of sodium vapours to their respective gaseous captions respectively are :

Options

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

Answer

To find the energy required to convert all the atoms present in 7 mg of lithium vapour and 23 mg of sodium vapour to their respective gaseous cations, we can follow these steps:

### Step 1: Calculate the number of moles of lithium (Li)
The molar mass of lithium (Li) is approximately 7 g/mol.

To convert 7 mg of lithium to grams:
\[ 7 \text{ mg} = 7 \times 10^{-3} \text{ g} \]

Now, calculate the number of moles of lithium:
\[ \text{Number of moles of Li} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{7 \times 10^{-3} \text{ g}}{7 \text{ g/mol}} = 0.001 \text{ mol} \]

### Step 2: Calculate the energy required for lithium
The ionization energy of lithium is given as 520 kJ/mol.

To find the energy required for 0.001 moles of lithium:
\[ \text{Energy for Li} = \text{Number of moles} \times \text{Ionization energy} = 0.001 \text{ mol} \times 520 \text{ kJ/mol} = 0.52 \text{ kJ} \]

### Step 3: Calculate the number of moles of sodium (Na)
The molar mass of sodium (Na) is approximately 23 g/mol.

To convert 23 mg of sodium to grams:
\[ 23 \text{ mg} = 23 \times 10^{-3} \text{ g} \]

Now, calculate the number of moles of sodium:
\[ \text{Number of moles of Na} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{23 \times 10^{-3} \text{ g}}{23 \text{ g/mol}} = 0.001 \text{ mol} \]

### Step 4: Calculate the energy required for sodium
The ionization energy of sodium is given as 495 kJ/mol.

To find the energy required for 0.001 moles of sodium:
\[ \text{Energy for Na} = \text{Number of moles} \times \text{Ionization energy} = 0.001 \text{ mol} \times 495 \text{ kJ/mol} = 0.495 \text{ kJ} \]

### Step 5: Total energy required
Now, we can sum the energies calculated for lithium and sodium:
\[ \text{Total Energy} = \text{Energy for Li} + \text{Energy for Na} = 0.52 \text{ kJ} + 0.495 \text{ kJ} = 1.015 \text{ kJ} \]

### Final Answer
The total energy required to convert all the atoms present in 7 mg of lithium vapour and 23 mg of sodium vapour to their respective gaseous cations is **1.015 kJ**.

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