What volume of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right),$ a nonelectrolyte,must be added to 15.0 \(\mathrm{L}\) water to produce an antifreeze solution with a freezing point of \(-25.0^{\circ} \mathrm{C} ?\) What is the boiling point of this solution? (The density of ethylene glycol is \(1.11 \mathrm{g} / \mathrm{cm}^{3},\) and the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3} . )\)

To find the volume of ethylene glycol needed to produce an antifreeze solution with a freezing point of -25.0°C, we need to first find the molality, then convert it to moles and finally to volume. The molality is calculated as: \(m = \frac{25.0}{1.86}\). This gives us the moles of ethylene glycol needed: \(Moles_{ethylene\ glycol} = \frac{25.0}{1.86} \times \frac{15000}{1000} \). Then, we can find the mass of ethylene glycol and its volume: \(Volume_{ethylene\ glycol} = \frac{Mass_{ethylene\ glycol}}{Density_{ethylene\ glycol}}\). To find the boiling point of this solution, we can use the boiling point elevation formula: \(\Delta T_b = 1 \cdot 0.512 \cdot \frac{25.0}{1.86}\). The final boiling point would be the sum of the boiling point elevation and the normal boiling point of water (100°C): \(Boiling\ point_{solution} = 100 + \Delta T_b\).