A solid cube of edge length =25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water.The boiling point of this solution is found to be `100.52^@C`(assume 100% ionisation of ionic compound).If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer).
(`K_b` of water =0.52 K kg `"mole"^(-1)`,Avogadro's number,`N_A=6xx10^(23),(root3(75))=4.22`)

Explanation

To solve the problem step by step, we will follow the method outlined in the video transcript and apply the relevant formulas.

### Step 1: Calculate the change in boiling point (ΔT)
The boiling point of the solution is given as 100.52 °C, and the boiling point of pure water is 100 °C. Therefore, the change in boiling point (ΔT) is:

\[ \Delta T = 100.52 °C - 100 °C = 0.52 °C \]

### Step 2: Use the boiling point elevation formula
The boiling point elevation can be calculated using the formula:

\[ \Delta T = K_b \cdot m \]

Where:
- \( K_b \) is the ebullioscopic constant of water (0.52 K kg/mol)
- \( m \) is the molality of the solution

Rearranging the formula to find molality:

\[ m = \frac{\Delta T}{K_b} = \frac{0.52}{0.52} = 1 \text{ mol/kg} \]

### Step 3: Calculate the number of moles of ionic compound
Since the molality is 1 mol/kg and we have 1 kg of water, the number of moles of the ionic compound is:

\[ \text{Number of moles} = 1 \text{ mol} \]

### Step 4: Calculate the total number of ions
Assuming 100% ionization, NaCl dissociates into Na\(^+\) and Cl\(^-\). Therefore, for every mole of NaCl, we get 2 moles of ions:

\[ \text{Total ions} = 2 \times \text{Number of moles} = 2 \times 1 = 2 \text{ moles of ions} \]

### Step 5: Calculate the total number of ions in terms of particles
Using Avogadro's number (\( N_A = 6 \times 10^{23} \) particles/mol):

\[ \text{Total number of ions} = 2 \text{ moles} \times 6 \times 10^{23} \text{ particles/mol} = 1.2 \times 10^{24} \text{ ions} \]

### Step 6: Calculate the number of unit cells
The number of unit cells can be calculated as follows:

\[ \text{Number of unit cells} = \frac{\text{Total number of ions}}{4} = \frac{1.2 \times 10^{24}}{4} = 3.0 \times 10^{23} \text{ unit cells} \]

### Step 7: Calculate the number of unit cells along one edge of the cube
The edge length of the cube is given as 25.32 mm, which is:

\[ \text{Edge length} = 25.32 \times 10^{-3} \text{ m} = 25.32 \times 10^{-3} \text{ m} = 25.32 \times 10^{6} \text{ pm} \]

Now, the number of unit cells along one edge of the cube is:

\[ \text{Number of unit cells along one edge} = \sqrt[3]{3.0 \times 10^{23}} \approx 6.7 \times 10^{7} \]

### Step 8: Calculate the edge length of one unit cell
The edge length of one unit cell (a) can be calculated as:

\[ a = \frac{\text{Edge length of cube}}{\text{Number of unit cells along one edge}} = \frac{25.32 \times 10^{6} \text{ pm}}{6.7 \times 10^{7}} \approx 600 \text{ pm} \]

### Step 9: Relate the edge length to the radii of ions
For NaCl type lattice, the relationship between the edge length (a) and the ionic radii is given by:

\[ a = 2(r_{cation} + r_{anion}) \]

Given that the radius of the anion \( r_{anion} = 200 \text{ pm} \):

\[ 600 = 2(r_{cation} + 200) \]

### Step 10: Solve for the radius of the cation
Rearranging the equation:

\[ 600 = 2r_{cation} + 400 \]

\[ 2r_{cation} = 600 - 400 = 200 \]

\[ r_{cation} = \frac{200}{2} = 100 \text{ pm} \]

### Final Answer
The radius of the cation of the solid is **100 pm**.

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