91Ó°ÊÓ

Write down the information given in the problem: Mass of lysozyme: \(20.0\mathrm{~mg}\) Volume of solution: \(225\mathrm{~mL}\) Temperature: \(T = 23^{\circ}\mathrm{C} = 23 + 273.15 = 296.15\mathrm{~K}\) Osmotic pressure: \(\Pi = 0.118\mathrm{~mm~Hg}\)

To use the osmotic pressure in calculations, it needs to be in atm. Convert osmotic pressure from mm Hg to atm using the conversion factor 1 atm = 760 mm Hg: \(\Pi = 0.118\mathrm{~mm~Hg} \times \frac{1\mathrm{~atm}}{760\mathrm{~mm~Hg}} = 1.553\times10^{-4}\mathrm{~atm}\)

The mass of lysozyme needs to be in grams for our calculations. Mass of lysozyme: \(20.0\mathrm{~mg} \times \frac{1\mathrm{~g}}{1000\mathrm{~mg}} = 0.020\mathrm{~g}\)

The formula to use for osmotic pressure, \(\Pi\), is given by: \(\Pi = CRT\) Where: \(C\) is the molar concentration of the solute, \(R\) is the ideal gas constant (\(0.08206 \mathrm{~L~atm~mol^{-1}~K^{-1}}\)), \(T\) is the temperature in Kelvin. First, we need to find the concentration \(C\) with the formula: \(C = \frac{n}{V}\) Where \(n\) is the number of moles of solute and \(V\) is the volume of the solution in liters. The number of moles, \(n\), can be expressed as the mass of the solute, \(m\), divided by its molar mass, \(M\). Therefore, we have the concentration as: \(C = \frac{m}{MV}\) Now, we can substitute this expression for \(C\) into the osmotic pressure formula: \(\Pi = \frac{mRT}{MV}\) We want to find the molar mass \(M\), so we can rearrange this equation for \(M\): \(M = \frac{mRT}{\Pi V}\) Now plug in the values we found earlier: \(M = \frac{0.020\mathrm{~g} \times 0.08206\mathrm{~L~atm~mol^{-1}~K^{-1}} \times 296.15\mathrm{~K}}{1.553\times10^{-4}\mathrm{~atm} \times 0.225\mathrm{~L}}\)

Calculate the molar mass of lysozyme: \(M = \frac{0.020\mathrm{~g} \times 0.08206\mathrm{~L~atm~mol^{-1}~K^{-1}} \times 296.15\mathrm{~K}}{1.553\times10^{-4}\mathrm{~atm} \times 0.225\mathrm{~L}} \approx 24310\mathrm{~g~mol^{-1}}\) The estimated molar mass of lysozyme is around \(24,310 \mathrm{~g~mol^{-1}}\).