What is the boiling point of 1 molal aqueous solution of NaCl `(K_b)=0.52 K molal^-1]`

Options

A. `99.48^@C`
B. `98.96^@C`
C. `100.52^@C`
D. `101.04^@C`

Answer

To find the boiling point of a 1 molal aqueous solution of NaCl, we can use the formula for boiling point elevation:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:
- \(\Delta T_b\) = boiling point elevation
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(K_b\) = ebullioscopic constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 1: Identify the values
- Given:
- \(K_b = 0.52 \, \text{K kg mol}^{-1}\)
- \(m = 1 \, \text{molal}\)

### Step 2: Determine the van 't Hoff factor (i)
For NaCl, it dissociates into two ions:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
Thus, \(i = 2\) because one formula unit of NaCl produces two ions.

### Step 3: Calculate the boiling point elevation (\(\Delta T_b\))
Substituting the values into the boiling point elevation formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
\[ \Delta T_b = 2 \cdot 0.52 \, \text{K kg mol}^{-1} \cdot 1 \, \text{molal} \]
\[ \Delta T_b = 1.04 \, \text{K} \]

### Step 4: Calculate the new boiling point of the solution
The boiling point of pure water is \(100 \, \text{°C}\). Therefore, the boiling point of the solution is:
\[ T_b = T_b(\text{solvent}) + \Delta T_b \]
\[ T_b = 100 \, \text{°C} + 1.04 \, \text{K} \]
\[ T_b = 101.04 \, \text{°C} \]

### Final Answer
The boiling point of the 1 molal aqueous solution of NaCl is **101.04 °C**.

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