Question

1. 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 0.0545 atm at 25.0 °c. calculate the molar mass of the protein. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall the osmotic pressure formula

The osmotic pressure formula is \(\Pi = MRT\), where \(\Pi\) is the osmotic pressure, \(M\) is the molarity, \(R\) is the gas constant (\(R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}}\)), and \(T\) is the absolute temperature. First, convert the temperature from Celsius to Kelvin: \(T = 25.0 + 273.15 = 298.15\ \text{K}\). The volume of the solution is \(V = 5.00\ \text{mL} = 0.00500\ \text{L}\). The mass of the protein is \(m = 247\ \text{mg} = 0.247\ \text{g}\).

Step2: Solve for molarity \(M\) from the osmotic pressure formula

Rearrange the formula to solve for \(M\): \(M=\frac{\Pi}{RT}\). Substitute the given values: \(\Pi = 0.0545\ \text{atm}\), \(R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}}\), \(T = 298.15\ \text{K}\). So \(M=\frac{0.0545\ \text{atm}}{0.0821\ \frac{\text{L·atm}}{\text{mol·K}} \times 298.15\ \text{K}}\). Calculate the denominator: \(0.0821\times298.15\approx24.48\). Then \(M=\frac{0.0545}{24.48}\approx0.002226\ \text{mol/L}\).

Step3: Calculate the number of moles \(n\) of the protein

Molarity \(M=\frac{n}{V}\), so \(n = M\times V\). Substitute \(M = 0.002226\ \text{mol/L}\) and \(V = 0.00500\ \text{L}\): \(n = 0.002226\ \text{mol/L} \times 0.00500\ \text{L}= 1.113\times 10^{-5}\ \text{mol}\).

Step4: Calculate the molar mass \(M_m\)

Molar mass is given by \(M_m=\frac{m}{n}\), where \(m = 0.247\ \text{g}\) and \(n = 1.113\times 10^{-5}\ \text{mol}\). So \(M_m=\frac{0.247\ \text{g}}{1.113\times 10^{-5}\ \text{mol}}\approx2.22\times 10^{4}\ \text{g/mol}\).

Answer:

\(2.22\times 10^{4}\ \frac{\text{g}}{\text{mol}}\)