117 g of NaCl is added to 222 g of water in a saucepan. At what does temperature does water boil at 101.325 kPa? Ebullioscopy constant for water = 0.52 K kg mol-1 and b.p. = 100°C

Options

A. 98.3°C
B. 102.8°C
C. 104.7°C
D. 101.5°C

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Mass of NaCl (solute), \( w_2 = 117 \, \text{g} \)
- Mass of water (solvent), \( w_1 = 222 \, \text{g} \)
- Boiling point of pure water, \( T_b = 100 \, \text{°C} \)
- Ebullioscopic constant for water, \( K_b = 0.52 \, \text{K kg mol}^{-1} \)
- Atmospheric pressure, \( P = 101.325 \, \text{kPa} \) (not directly needed for boiling point elevation calculation)

### Step 2: Calculate the number of moles of NaCl
To find the number of moles of NaCl, we need to use its molar mass. The molar mass of NaCl is:
- Sodium (Na) = 23 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molar mass of NaCl = \( 23 + 35.5 = 58.5 \, \text{g/mol} \)

Now, we can calculate the number of moles of NaCl:
\[ n_2 = \frac{w_2}{\text{Molar mass of NaCl}} = \frac{117 \, \text{g}}{58.5 \, \text{g/mol}} = 2 \, \text{mol} \]

### Step 3: Convert the mass of water to kilograms
Since the ebullioscopic constant is given in K kg mol\(^{-1}\), we need to convert the mass of water from grams to kilograms:
\[ w_1 = 222 \, \text{g} = 0.222 \, \text{kg} \]

### Step 4: Calculate the boiling point elevation (\( \Delta T_b \))
The formula for boiling point elevation is given by:
\[ \Delta T_b = K_b \cdot m_2 \]
where \( m_2 \) is the molality of the solution, calculated as:
\[ m_2 = \frac{n_2}{w_1 \, (\text{in kg})} = \frac{2 \, \text{mol}}{0.222 \, \text{kg}} \approx 9.01 \, \text{mol/kg} \]

Now substituting the values into the boiling point elevation formula:
\[ \Delta T_b = K_b \cdot m_2 = 0.52 \, \text{K kg mol}^{-1} \cdot 9.01 \, \text{mol/kg} \approx 4.69 \, \text{°C} \]

### Step 5: Calculate the new boiling point
The new boiling point of the solution is:
\[ T_{\text{new}} = T_b + \Delta T_b = 100 \, \text{°C} + 4.69 \, \text{°C} \approx 104.69 \, \text{°C} \]

### Final Answer
The boiling point of the water when 117 g of NaCl is added to 222 g of water is approximately **104.69 °C**.

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