Acetaminophen (C8H9NO2; MOLAR MASS 151.163 G/MOL) is the main ingredient of Tylenol. When 2.68 g acetaminophen dissolves in 45.3 g of solvent XYZ, the solution boils at 82.3 degrees celsius. If the boiling point of pure solvent XYZ (without acetaminophen) is 79.1 degrees Celcius, determine the boiling point elevation constant (in m/℃) of solvent XYZ. (Answer must have 3 SIGNIFICANT FIGURES).
The Correct Answer and Explanation is :
To solve this problem, we will use the formula for boiling point elevation:
[
\Delta T_b = K_b \times m
]
Where:
- (\Delta T_b) is the change in boiling point (in °C),
- (K_b) is the boiling point elevation constant of the solvent (in °C·kg/mol),
- (m) is the molality of the solution (in mol/kg).
Step 1: Calculate the change in boiling point ((\Delta T_b))
The boiling point elevation is the difference between the boiling point of the solution and the boiling point of the pure solvent:
[
\Delta T_b = T_b(\text{solution}) – T_b(\text{pure solvent})
]
Given:
- (T_b(\text{solution}) = 82.3 \, ^\circ \text{C}),
- (T_b(\text{pure solvent}) = 79.1 \, ^\circ \text{C}).
[
\Delta T_b = 82.3 \, ^\circ \text{C} – 79.1 \, ^\circ \text{C} = 3.2 \, ^\circ \text{C}
]
Step 2: Calculate the molality of the solution ((m))
Molality is defined as the number of moles of solute per kilogram of solvent:
[
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
]
To find the moles of acetaminophen (solute), we use its molar mass. Given:
- Mass of acetaminophen = 2.68 g,
- Molar mass of acetaminophen = 151.163 g/mol.
[
\text{moles of acetaminophen} = \frac{2.68 \, \text{g}}{151.163 \, \text{g/mol}} = 0.0177 \, \text{mol}
]
Now, we calculate the molality:
- Mass of solvent = 45.3 g = 0.0453 kg.
[
m = \frac{0.0177 \, \text{mol}}{0.0453 \, \text{kg}} = 0.391 \, \text{mol/kg}
]
Step 3: Solve for the boiling point elevation constant ((K_b))
Now we can solve for (K_b) using the equation for boiling point elevation:
[
\Delta T_b = K_b \times m
]
Rearranging:
[
K_b = \frac{\Delta T_b}{m}
]
Substitute the known values:
[
K_b = \frac{3.2 \, ^\circ \text{C}}{0.391 \, \text{mol/kg}} = 8.18 \, ^\circ \text{C} \cdot \text{kg/mol}
]
Final Answer:
The boiling point elevation constant ((K_b)) for solvent XYZ is 8.18 °C·kg/mol (rounded to 3 significant figures).
Now I will generate the image to show this process step by step.
Here is the image showing the step-by-step explanation of how to solve the boiling point elevation problem. Let me know if you need further clarification!
