Answer :
We start by noting that the boiling point of a solution can be determined by the boiling point elevation formula:
[tex]$$
\Delta T_b = i \cdot K_b \cdot m
$$[/tex]
where
- [tex]$i$[/tex] is the van 't Hoff factor (for sodium chloride, [tex]$i \approx 2$[/tex] because it dissociates into two ions),
- [tex]$K_b$[/tex] is the ebullioscopic constant for water (approximately [tex]$0.512\ ^\circ\text{C}\cdot\text{kg/mol}$[/tex]),
- [tex]$m$[/tex] is the molality of the solution in moles of solute per kilogram of solvent.
Step 1. Determine the mass of water and salt.
It is given that 1 cup of water weighs approximately 200 grams, and each tablespoon of salt weighs about 18 grams. With 6 tablespoons of salt, the total salt mass is:
[tex]$$
\text{Salt mass} = 6 \times 18 = 108 \text{ grams}
$$[/tex]
Step 2. Calculate the number of moles of salt.
The molar mass of sodium chloride (NaCl) is approximately 58.44 g/mol. Therefore, the number of moles of salt is computed as:
[tex]$$
\text{moles of salt} = \frac{108}{58.44} \approx 1.848 \text{ moles}
$$[/tex]
Step 3. Calculate the molality of the solution.
Molality is defined as the number of moles of solute per kilogram of solvent. Here, the mass of the solvent (water) is 200 grams, which is 0.2 kilograms. So,
[tex]$$
m = \frac{\text{moles of salt}}{\text{mass of water in kg}} = \frac{1.848}{0.2} \approx 9.240 \text{ mol/kg}
$$[/tex]
Step 4. Calculate the boiling point elevation.
Now, substitute into the boiling point elevation formula:
[tex]$$
\Delta T_b = i \cdot K_b \cdot m = 2 \times 0.512 \times 9.240 \approx 9.462\ ^\circ\text{C}
$$[/tex]
Step 5. Determine the new boiling point.
The normal boiling point of pure water is [tex]$100\ ^\circ\text{C}$[/tex]. With the elevation,
[tex]$$
\text{Boiling point} = 100 + \Delta T_b \approx 100 + 9.462 = 109.462\ ^\circ\text{C}
$$[/tex]
When rounded to the nearest given option, this value is approximately [tex]$110\ ^\circ\text{C}$[/tex].
Thus, the answer is:
[tex]$$
\boxed{110\ ^\circ\text{C}}
$$[/tex]
[tex]$$
\Delta T_b = i \cdot K_b \cdot m
$$[/tex]
where
- [tex]$i$[/tex] is the van 't Hoff factor (for sodium chloride, [tex]$i \approx 2$[/tex] because it dissociates into two ions),
- [tex]$K_b$[/tex] is the ebullioscopic constant for water (approximately [tex]$0.512\ ^\circ\text{C}\cdot\text{kg/mol}$[/tex]),
- [tex]$m$[/tex] is the molality of the solution in moles of solute per kilogram of solvent.
Step 1. Determine the mass of water and salt.
It is given that 1 cup of water weighs approximately 200 grams, and each tablespoon of salt weighs about 18 grams. With 6 tablespoons of salt, the total salt mass is:
[tex]$$
\text{Salt mass} = 6 \times 18 = 108 \text{ grams}
$$[/tex]
Step 2. Calculate the number of moles of salt.
The molar mass of sodium chloride (NaCl) is approximately 58.44 g/mol. Therefore, the number of moles of salt is computed as:
[tex]$$
\text{moles of salt} = \frac{108}{58.44} \approx 1.848 \text{ moles}
$$[/tex]
Step 3. Calculate the molality of the solution.
Molality is defined as the number of moles of solute per kilogram of solvent. Here, the mass of the solvent (water) is 200 grams, which is 0.2 kilograms. So,
[tex]$$
m = \frac{\text{moles of salt}}{\text{mass of water in kg}} = \frac{1.848}{0.2} \approx 9.240 \text{ mol/kg}
$$[/tex]
Step 4. Calculate the boiling point elevation.
Now, substitute into the boiling point elevation formula:
[tex]$$
\Delta T_b = i \cdot K_b \cdot m = 2 \times 0.512 \times 9.240 \approx 9.462\ ^\circ\text{C}
$$[/tex]
Step 5. Determine the new boiling point.
The normal boiling point of pure water is [tex]$100\ ^\circ\text{C}$[/tex]. With the elevation,
[tex]$$
\text{Boiling point} = 100 + \Delta T_b \approx 100 + 9.462 = 109.462\ ^\circ\text{C}
$$[/tex]
When rounded to the nearest given option, this value is approximately [tex]$110\ ^\circ\text{C}$[/tex].
Thus, the answer is:
[tex]$$
\boxed{110\ ^\circ\text{C}}
$$[/tex]
