288. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.0780 atm at 25.0 °C. Calculate the molar mass of the protein. Be sure your answer has the correct number of significant digits.
288. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.0780 atm at 25.0 °C. Calculate the molar mass of the protein. Be sure your answer has the correct number of significant digits.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Osmotic Pressure and Molar Mass of a Protein**
Given:
- 288 mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution.
- The osmotic pressure of this solution is measured to be 0.0780 atm at 25.0 °C.
**Objective:**
Calculate the molar mass of the protein. Ensure your answer has the correct number of significant digits.
1. **Formula Involved:**
\[
\Pi = MRT
\]
where:
- \(\Pi\) = osmotic pressure (in atm)
- \(M\) = molarity (in mol/L)
- \(R\) = ideal gas constant \(0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1}\)
- \(T\) = temperature (in Kelvin)
2. **Given Data:**
- \(\Pi = 0.0780 \, \text{atm}\)
- Volume of solution V = 5.00 mL = 0.00500 L
- Temperature \( T = 25.0 °C = 298.15 \, K \)
3. **Conversion:**
- Mass of protein = 288 mg = 0.288 g
**Calculation:**
- First, calculate the molarity \( M \) of the solution:
\[
M = \frac{\Pi}{RT} = \frac{0.0780 \, \text{atm}}{(0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1} \times 298.15 \, K)}
\]
- Using the values from above:
\[
M = \frac{0.0780}{24.466215} \, \text{mol/L} \approx 0.00319 \, \text{mol/L}
\]
- Find the number of moles of protein in 5.00 mL (0.005 L) of solution:
\[
n = M \times V = 0.00319 \, \text{mol/L} \times 0.005 \, \text{L} \approx 1.595 \times 10^{-5}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F688553c1-a29d-4d73-a765-adb59e2cd2b6%2Fb49700f3-46a7-4b43-b605-ce120f804c5e%2Fuefzn8_processed.png&w=3840&q=75)
Transcribed Image Text:**Osmotic Pressure and Molar Mass of a Protein**
Given:
- 288 mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution.
- The osmotic pressure of this solution is measured to be 0.0780 atm at 25.0 °C.
**Objective:**
Calculate the molar mass of the protein. Ensure your answer has the correct number of significant digits.
1. **Formula Involved:**
\[
\Pi = MRT
\]
where:
- \(\Pi\) = osmotic pressure (in atm)
- \(M\) = molarity (in mol/L)
- \(R\) = ideal gas constant \(0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1}\)
- \(T\) = temperature (in Kelvin)
2. **Given Data:**
- \(\Pi = 0.0780 \, \text{atm}\)
- Volume of solution V = 5.00 mL = 0.00500 L
- Temperature \( T = 25.0 °C = 298.15 \, K \)
3. **Conversion:**
- Mass of protein = 288 mg = 0.288 g
**Calculation:**
- First, calculate the molarity \( M \) of the solution:
\[
M = \frac{\Pi}{RT} = \frac{0.0780 \, \text{atm}}{(0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1} \times 298.15 \, K)}
\]
- Using the values from above:
\[
M = \frac{0.0780}{24.466215} \, \text{mol/L} \approx 0.00319 \, \text{mol/L}
\]
- Find the number of moles of protein in 5.00 mL (0.005 L) of solution:
\[
n = M \times V = 0.00319 \, \text{mol/L} \times 0.005 \, \text{L} \approx 1.595 \times 10^{-5}
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