At \(25^{\circ} \mathrm{C}\) a \(0.50 \mathrm{g}\) sample of polyisobutylene (a polymer used in synthetic rubber) in \(100.0 \mathrm{mL}\) of benzene solution has an osmotic pressure that supports a \(5.1 \mathrm{mm}\) column of solution \((d=0.88 \mathrm{g} / \mathrm{mL}) .\) What is the molar mass of the polyisobutylene? (For \(\mathrm{Hg}\), \(d=13.6 \mathrm{g} / \mathrm{mL} .)\)

To convert from the height of a column to pressure, we use the formula \[P = d \cdot g \cdot h\], where \(P\) is the pressure, \(d\) is the density, \(g\) is the acceleration due to gravity, and \(h\) is the height. We also need to convert the density of solution from \(\mathrm{g/mL}\) to \(\mathrm{kg/m^3}\) (standard SI units), the height from mm to m, and the pressure to atmosphere (Atm) since it is the unit used in the ideal gas law. As the density of the liquid column is given in terms of the solution and the unit weight of mercury is also provided, the pressure \(P\) in terms of mercury can be given as \(P = d_{solution}/d_{mercury} \cdot h_{solution}\). Here, \(d_{solution} = 0.88 g/mL = 880 kg/m^3\), \(d_{mercury} = 13.6 g/mL = 13600 kg/m^3\), and \(h_{solution} = 5.1 mm = 0.0051 m\). The acceleration due to gravity \(g\) is approximately \(9.8 m/s^2\). So the pressure \(P = (880/13600) \cdot 0.0051 \cdot 9.8 m/s^2 = 0.033 Atm \)

The osmotic pressure equation \[ \pi = \frac{n}{V}RT \] , can be rewritten as \[ \pi = \frac{m}{M} \cdot \frac{RT}{V} \] , where \(m\) is the mass of the solute, \(M\) is the molar mass of the solute, \(R\) is the ideal gas constant, \(T\) is the temperature, and \(V\) is the volume of solution. Panel The volume \(V = 100.0 mL = 0.1 L\), the temperature \(T = 25^{\circ}C = 298.15 K\), and \(R = 0.0821 \,L \cdot Atm/mol \cdot K \) (the value of the ideal gas constant).

Rearranging the osmotic pressure equation for \(M\) gives us \( M = \frac{m \cdot RT}{\pi \cdot V} \). Substituting the known values gives \( M = \frac{(0.50 g) \cdot (0.0821 L \cdot Atm/mol \cdot K) \cdot (298.15 K)}{ (0.033 Atm) \cdot (0.1 L)} = 37.43 g/mol \).