A 300 mg sample of caffeine was dissolved in 10.0 g of camphor (Ke=39.7°C/m), decreasing the freezing point of camphor by 3.07°C. What is the molar mass of caffeine? 1.00 kg = 1000g 47.0 g/mol O a. 388 g/mol Ob. 97.0 g/mol Oc. 194 g/mol d. 94.0 g/mol Oe. Question 2
A 300 mg sample of caffeine was dissolved in 10.0 g of camphor (Ke=39.7°C/m), decreasing the freezing point of camphor by 3.07°C. What is the molar mass of caffeine? 1.00 kg = 1000g 47.0 g/mol O a. 388 g/mol Ob. 97.0 g/mol Oc. 194 g/mol d. 94.0 g/mol Oe. Question 2
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Question:**
A 300 mg sample of caffeine was dissolved in 10.0 g of camphor (\(K_f = 39.7 \degree \text{C/m}\)), decreasing the freezing point of camphor by \( 3.07 \degree \text{C} \). What is the molar mass of caffeine?
1.00 kg = 1000 g
- a. 47.0 g/mol
- b. 388 g/mol
- c. 97.0 g/mol
- d. 194 g/mol
- e. 94.0 g/mol
**Explanation:**
To determine the molar mass of caffeine, we utilize the formula for freezing point depression:
\[ \Delta T_f = K_f \cdot m \]
where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution.
Given:
- \( \Delta T_f = 3.07 \degree \text{C} \)
- \( K_f = 39.7 \degree \text{C/m} \)
- Mass of camphor (solvent) = 10.0 g = 0.010 kg
- Mass of caffeine (solute) = 300 mg = 0.300 g
First, we calculate the molality (\( m \)) of the solution:
\[ m = \frac{\Delta T_f}{K_f} = \frac{3.07 \degree \text{C}}{39.7 \degree \text{C/m}} = 0.0773 \text{ m} \]
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent, so:
\[ m = \frac{\text{moles of caffeine}}{\text{kg of camphor}} \]
We rearrange to find the moles of caffeine:
\[ \text{moles of caffeine} = m \times \text{kg of camphor} = 0.0773 \text{ m} \times 0.010 \text{ kg} = 0.000773 \text{ moles} \]
Finally, we find the molar mass of caffeine (\( M \)) using:
\[ M = \frac{\text{mass of caffeine](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e0b628-1887-4036-8cfa-486903c53c1e%2F27403100-7020-4264-b5fd-33aaa785effb%2Fl588q5k9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
A 300 mg sample of caffeine was dissolved in 10.0 g of camphor (\(K_f = 39.7 \degree \text{C/m}\)), decreasing the freezing point of camphor by \( 3.07 \degree \text{C} \). What is the molar mass of caffeine?
1.00 kg = 1000 g
- a. 47.0 g/mol
- b. 388 g/mol
- c. 97.0 g/mol
- d. 194 g/mol
- e. 94.0 g/mol
**Explanation:**
To determine the molar mass of caffeine, we utilize the formula for freezing point depression:
\[ \Delta T_f = K_f \cdot m \]
where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution.
Given:
- \( \Delta T_f = 3.07 \degree \text{C} \)
- \( K_f = 39.7 \degree \text{C/m} \)
- Mass of camphor (solvent) = 10.0 g = 0.010 kg
- Mass of caffeine (solute) = 300 mg = 0.300 g
First, we calculate the molality (\( m \)) of the solution:
\[ m = \frac{\Delta T_f}{K_f} = \frac{3.07 \degree \text{C}}{39.7 \degree \text{C/m}} = 0.0773 \text{ m} \]
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent, so:
\[ m = \frac{\text{moles of caffeine}}{\text{kg of camphor}} \]
We rearrange to find the moles of caffeine:
\[ \text{moles of caffeine} = m \times \text{kg of camphor} = 0.0773 \text{ m} \times 0.010 \text{ kg} = 0.000773 \text{ moles} \]
Finally, we find the molar mass of caffeine (\( M \)) using:
\[ M = \frac{\text{mass of caffeine
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