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 \( 25.0^{\circ} \mathrm{C} \). Calculate the molar mass of the protein. Be sure your answer has the correct number of significant digits. \( \square \frac{\mathrm{g}}{\mathrm{mol}} \)

To calculate the molar mass of the protein using osmotic pressure, you can use the formula: \[ \pi = \frac{n}{V}RT \] Where: - \( \pi \) is the osmotic pressure (0.0594 atm) - \( n \) is the number of moles of solute - \( V \) is the volume of the solution in liters (5.00 mL = 0.005 L) - \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)) - \( T \) is the temperature in Kelvin (25.0 °C = 298.15 K) First, rearranging the equation to find \( n \): \[ n = \frac{\pi V}{RT} \] Substituting the known values: \[ n = \frac{0.0594 \, \text{atm} \times 0.005 \, \text{L}}{0.0821 \, \text{L·atm/(K·mol)} \times 298.15 \, \text{K}} \approx 0.00012 \, \text{mol} \] Next, to find the molar mass (\(M\)), use the equation: \[ M = \frac{mass}{n} \] The mass of the protein in grams is 273 mg, which is 0.273 g. \[ M = \frac{0.273 \, \text{g}}{0.00012 \, \text{mol}} \approx 2275 \, \text{g/mol} \] Rounding to three significant figures (since 273 mg has three significant figures), the molar mass of the protein is: \[ \boxed{2270 \, \frac{\text{g}}{\text{mol}}} \]