A 0.500 -g sample of a compound is dissolved in enough water to form \(100.0 \mathrm{mL}\) of solution. This solution has an osmotic pressure of 2.50 atm at \(25^{\circ} \mathrm{C}\). If each molecule of the solute dissociates into two particles (in this solvent), what is the molar mass of this solute?

To work with the osmotic pressure formula, we need to convert the given temperature from Celsius to Kelvin: \[ T_{K} = T_{C} + 273.15\] \[T_{K} = 25^{\circ}C + 273.15 = 298.15\,\mathrm{K}\]

Using the osmotic pressure formula, we can rearrange it to solve for the moles of solute particles: \[\cfrac{n}{V} = \cfrac{\Pi}{R \cdot T}\] We are given the osmotic pressure \(\Pi = 2.50\,\mathrm{atm}\), the volume of the solution \(V = 100.0\,\mathrm{mL} = 0.100\,\mathrm{L}\), and the ideal gas constant \(R = 0.0821\,\cfrac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}\). Now we can plug the values into the formula: \[\cfrac{n}{0.100\,\mathrm{L}} = \cfrac{2.50\,\mathrm{atm}}{(0.0821\,\cfrac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}) \cdot (298.15\,\mathrm{K})}\] From this, we can solve for the moles of solute particles, \(n\): \[n = 0.100\,\mathrm{L} \cdot \cfrac{2.50\,\mathrm{atm}}{(0.0821\,\cfrac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}) \cdot (298.15\,\mathrm{K})} = 0.0102\,\mathrm{mol}\]

Since each molecule of the solute dissociates into two particles in the solvent, there are half as many moles of the solute compound as there are solute particles: \[n_{\mathrm{solute}} = \cfrac{n_{\mathrm{particles}}}{2}\] \[n_{\mathrm{solute}} = \cfrac{0.0102\,\mathrm{mol}}{2} = 0.0051\,\mathrm{mol}\]

We are given the sample mass as 0.500 g. Now we can find the molar mass of the solute using the following equation: \[M_{\mathrm{solute}} = \cfrac{m_{\mathrm{solute}}}{n_{\mathrm{solute}}}\] \[M_{\mathrm{solute}} = \cfrac{0.500\,\mathrm{g}}{0.0051\,\mathrm{mol}} = 97.8\,\cfrac{\mathrm{g}}{\mathrm{mol}}\] The molar mass of the solute compound is approximately 97.8 g/mol.