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 0.127 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 osmotic pressure, \(M\) is molarity, \(R\) is the gas constant (\(0.0821\ \frac{\text{L·atm}}{\text{mol·K}}\)), and \(T\) is temperature in Kelvin. First, convert temperature to Kelvin: \(T = 25.0 + 273.15 = 298.15\ \text{K}\). Convert volume to liters: \(V = 5.00\ \text{mL} = 0.00500\ \text{L}\). Mass of protein: \(m = 168.\ \text{mg} = 0.168\ \text{g}\).

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

Rearrange \(\Pi = MRT\) to \(M=\frac{\Pi}{RT}\). Substitute values: \(\Pi = 0.127\ \text{atm}\), \(R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}}\), \(T = 298.15\ \text{K}\). So \(M=\frac{0.127}{0.0821\times298.15}\). Calculate denominator: \(0.0821\times298.15\approx24.48\). Then \(M\approx\frac{0.127}{24.48}\approx0.00519\ \text{mol/L}\).

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

Molarity \(M=\frac{n}{V}\), so \(n = M\times V\). Substitute \(M = 0.00519\ \text{mol/L}\) and \(V = 0.00500\ \text{L}\): \(n = 0.00519\times0.00500\approx2.595\times10^{-5}\ \text{mol}\).

Step4: Calculate molar mass (\(MM\))

Molar mass \(MM=\frac{m}{n}\), where \(m = 0.168\ \text{g}\) and \(n = 2.595\times10^{-5}\ \text{mol}\). So \(MM=\frac{0.168}{2.595\times10^{-5}}\approx6.47\times10^{3}\ \frac{\text{g}}{\text{mol}}\).

Answer:

\(\boldsymbol{6.5\times10^{3}\ \frac{\text{g}}{\text{mol}}}\) (or more precisely \(6.47\times10^{3}\ \frac{\text{g}}{\text{mol}}\) depending on significant figures; the given values have 3 significant figures for mass (168. mg), 4 for pressure (0.127 atm), 3 for volume (5.00 mL), 3 for temperature (25.0 °C), so the answer should have 3 significant figures, so \(6.47\times10^{3}\approx6.47\times10^{3}\) or rounded to \(6.5\times10^{3}\) if considering the calculation steps more precisely, but the exact calculation gives approximately \(6470\ \frac{\text{g}}{\text{mol}}\) or \(6.47\times10^{3}\ \frac{\text{g}}{\text{mol}}\))