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)`

Options

A. (a) 2.34 g
B. (b) 1.76 g
C. (c) 4.12 g
D. (d) 0.65 g

Explanation

To solve the problem of determining the amount of NaCl to be dissolved in 600 g of H₂O to decrease the freezing point by 0.2°C, we will use the formula for depression in freezing point, which is a colligative property. Here are the steps to arrive at the solution:

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

### Step 2: Identify the Given Values
From the problem, we have:
- \(\Delta T_f = 0.2°C\)
- \(K_f\) for H₂O = 2 °C kg/mol
- Mass of H₂O = 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⁻ ions, the van 't Hoff factor \(i\) is:
\[ i = 2 \]

### Step 4: Rearrange the Formula to Find Molality (m)
Rearranging the formula to find molality:
\[ m = \frac{\Delta T_f}{i \cdot K_f} \]
Substituting the known values:
\[ m = \frac{0.2}{2 \cdot 2} = \frac{0.2}{4} = 0.05 \text{ mol/kg} \]

### Step 5: Calculate the Number of Moles of NaCl Required
Using the definition of molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Let \(n\) be the number of moles of NaCl. Thus:
\[ 0.05 = \frac{n}{0.6} \]
Solving for \(n\):
\[ n = 0.05 \cdot 0.6 = 0.03 \text{ moles} \]

### Step 6: Calculate the Mass of NaCl Required
To find the mass of NaCl, we use its molar mass:
- Molar mass of NaCl = 58.5 g/mol

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

### Step 7: Round Off the Answer
Rounding off the answer, we get:
\[ \text{mass of NaCl} \approx 1.76 \text{ g} \]

### Final Answer
The amount of NaCl to be dissolved in 600 g of H₂O to decrease the freezing point by 0.2°C is approximately **1.76 g**.

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