P ALEKS1.3 - Spring 2025 Ceneral chemistry II (CHM-1220-01)
Using osmotic pressure to find molar mass
273. 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 0.0594 atm at .
Calculate the molar mass of the protein. Be sure your answer has the correct number of significant digits.

Ask by Washington Bond. in the United States

Jan 20,2025

Question

P ALEKS1.3 - Spring 2025 Ceneral chemistry II (CHM-1220-01)
Using osmotic pressure to find molar mass
273. 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 0.0594 atm at .
Calculate the molar mass of the protein. Be sure your answer has the correct number of significant digits.

Ask by Washington Bond. in the United States

Jan 20,2025

UpStudy AI · Accepted Answer

To calculate the molar mass of the unknown protein using osmotic pressure, we'll follow these steps:

### **Given Data:**
- **Mass of protein (m):** 273 mg = 0.273 g
- **Volume of solution (V):** 5.00 mL = 0.00500 L
- **Osmotic pressure (π):** 0.0594 atm
- **Temperature (T):** $$25.0^{\circ} \mathrm{C}$$ = 298 K
- **Gas constant (R):** 0.08206 L·atm/mol·K

### **Relevant Equation:**
The osmotic pressure ($$\pi$$) is related to the molar mass (M) by the equation:

$$
\pi = \frac{mRT}{MV}
$$

Solving for $$M$$:

$$
M = \frac{mRT}{\pi V}
$$

### **Plugging in the Values:**
$$
M = \frac{0.273 \, \text{g} \times 0.08206 \, \text{L·atm/mol·K} \times 298 \, \text{K}}{0.0594 \, \text{atm} \times 0.00500 \, \text{L}}
$$

### **Calculations:**
1. **Calculate the numerator:**
   $$
   0.273 \times 0.08206 \times 298 \approx 6.679 \, \text{g·L·atm/mol}
   $$
   
2. **Calculate the denominator:**
   $$
   0.0594 \times 0.00500 = 0.000297 \, \text{atm·L}
   $$
   
3. **Calculate $$M$$:**
   $$
   M = \frac{6.679}{0.000297} \approx 22,500 \, \text{g/mol}
   $$

### **Consider Significant Figures:**
All given values have **3 significant figures**, so the final answer should also be reported with **3 significant figures**.

### **Final Answer:**
$$
M = 2.25 \times 10^4 \, \frac{\mathrm{g}}{\mathrm{mol}}
$$

**Thus, the molar mass of the protein is $$2.25 \times 10^4 \, \mathrm{g/mol}$$.**
The molar mass of the protein is $$2.25 \times 10^4 \, \frac{\mathrm{g}}{\mathrm{mol}}$$.

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