Determine the amount of NaCl to be dissolved in 600g `H_(2)O` to decrease the freezing point by `0.2^@ C`.
Given : `k_f` of `H_(2)O = 2 km^(-1)` .
Density of `H_(2)O(l) = 1 g/ml`

Answer

To determine the amount of NaCl to be dissolved in 600 g of water to decrease the freezing point by 0.2°C, we can follow these steps:

### Step 1: Understand the Depression in Freezing Point Formula
The depression in freezing point (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) = depression in freezing point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_f \) = cryoscopic constant (freezing point depression constant) for the solvent
- \( m \) = molality of the solution

### Step 2: Identify Given Values
From the question:
- \( \Delta T_f = 0.2 \, °C \)
- \( K_f = 2 \, \text{K kg}^{-1} \)
- Mass of water (solvent) = 600 g = 0.6 kg (since 1 kg = 1000 g)

### Step 3: Determine the van 't Hoff Factor (i)
For NaCl, which dissociates into Na\(^+\) and Cl\(^-\):
- \( i = 2 \) (since it dissociates into 2 ions)

### Step 4: Substitute Values into the Depression Formula
Now, substitute the known values into the depression formula:
\[ 0.2 = 2 \cdot 2 \cdot m \]
This simplifies to:
\[ 0.2 = 4m \]

### Step 5: Solve for Molality (m)
Rearranging the equation to find \( m \):
\[ m = \frac{0.2}{4} = 0.05 \, \text{mol/kg} \]

### Step 6: Relate Molality to Moles of Solute
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Let \( n \) be the moles of NaCl. We can express this as:
\[ 0.05 = \frac{n}{0.6} \]

### Step 7: Solve for Moles of NaCl (n)
Rearranging gives:
\[ n = 0.05 \times 0.6 = 0.03 \, \text{moles} \]

### Step 8: Calculate the Mass of NaCl
To find the mass of NaCl, we use its molar mass:
- Molar mass of NaCl = 23 (Na) + 35.5 (Cl) = 58.5 g/mol

Now, calculate the mass:
\[ \text{mass of NaCl} = n \times \text{molar mass} = 0.03 \, \text{moles} \times 58.5 \, \text{g/mol} = 1.755 \, \text{g} \]

### Final Answer
The amount of NaCl to be dissolved is **1.755 g**.

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