The freezing point of `1` molal `NaCl` solution assuming `NaCl` to be `100%` dissociated in water is:

Options

A. `-1.86^(@)C`
B. `-3.72^(@)C`
C. `+1.86^(@)C`
D. `+3.72^(@)C`

Explanation

To find the freezing point of a 1 molal NaCl solution, we will follow these steps:

### Step 1: Understand the concept of freezing point depression
When a non-volatile solute like NaCl is dissolved in water, it lowers the freezing point of the solution compared to pure water. This phenomenon is known as freezing point depression.

### Step 2: Identify the formula for freezing point depression
The freezing point depression can be calculated using 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 of the solvent (for water, \(K_f = 1.86 \, \text{°C kg/mol}\))
- \(m\) = molality of the solution (in this case, \(1 \, \text{molal}\))

### Step 3: Determine the van 't Hoff factor (\(i\))
For NaCl, which dissociates completely in water:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
This means NaCl produces 2 ions in solution. Therefore, the van 't Hoff factor \(i\) is:
\[ i = 2 \]

### Step 4: Substitute the values into the formula
Now we can substitute the values into the freezing point depression formula:
\[ \Delta T_f = i \cdot K_f \cdot m = 2 \cdot 1.86 \, \text{°C kg/mol} \cdot 1 \, \text{molal} \]
Calculating this gives:
\[ \Delta T_f = 2 \cdot 1.86 = 3.72 \, \text{°C} \]

### Step 5: Calculate the new freezing point of the solution
The freezing point of pure water is \(0 \, \text{°C}\). To find the freezing point of the solution, we subtract the depression from the freezing point of pure water:
\[ T_f' = T_f - \Delta T_f = 0 \, \text{°C} - 3.72 \, \text{°C} = -3.72 \, \text{°C} \]

### Final Answer
The freezing point of the 1 molal NaCl solution is:
\[ \boxed{-3.72 \, \text{°C}} \]

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