At a certain Hill station, water boils at `96^(@)C`. The amount of NaCl that should be added to one litre of water so that it boils at `100^(@)C` will be (`K_(b)` for `H_(2)O=0.52K//m`)

Options

A. 450 g
B. 225 g
C. 125 g
D. 250 g

Explanation

To solve the problem of how much NaCl should be added to 1 liter of water to raise its boiling point from 96°C to 100°C, we can follow these steps:

### Step 1: Calculate the change in boiling point (ΔT_b)
The change in boiling point (ΔT_b) is calculated as follows:
\[ \Delta T_b = T_{final} - T_{initial} = 100°C - 96°C = 4°C \]

### Step 2: Use the boiling point elevation formula
The formula for boiling point elevation is:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \(i\) = van 't Hoff factor (for NaCl, \(i = 2\) because it dissociates into Na⁺ and Cl⁻)
- \(K_b\) = ebullioscopic constant of the solvent (for water, \(K_b = 0.52 \, K/m\))
- \(m\) = molality of the solution

### Step 3: Substitute known values into the formula
Substituting the known values into the formula:
\[ 4°C = 2 \cdot 0.52 \cdot m \]

### Step 4: Solve for molality (m)
Rearranging the equation to solve for molality:
\[ m = \frac{4°C}{2 \cdot 0.52} = \frac{4}{1.04} \approx 3.846 \, mol/kg \]

### Step 5: Calculate the number of moles of NaCl required
Since we are using 1 liter of water, which has a mass of 1 kg, the molality (m) is equal to the number of moles of solute (n) in 1 kg of solvent:
\[ n = m \cdot \text{mass of solvent in kg} = 3.846 \, mol/kg \cdot 1 \, kg = 3.846 \, mol \]

### Step 6: Calculate the mass of NaCl required
To find the mass of NaCl (W2), we use the formula:
\[ W_2 = n \cdot M \]
Where \(M\) is the molar mass of NaCl (approximately 58.5 g/mol):
\[ W_2 = 3.846 \, mol \cdot 58.5 \, g/mol \approx 225 \, g \]

### Final Answer
The amount of NaCl that should be added to 1 liter of water to raise its boiling point to 100°C is approximately **225 grams**.

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