91Ó°ÊÓ

The freezing-point depression ΔTf can be mathematically expressed as: \[ \Delta T_\text{f} = K_\text{f} \cdot m \cdot i \] where \(K_f\) is the cryoscopic constant of the solvent (in this case, water), \(m\) is the molality of the solute (ethylene glycol), and \(i\) is the van't Hoff factor, which represents the number of particles the solute forms in the solution. For ethylene glycol, \(i\) equals to 1 because it does not dissociate into ions in the solution. The cryoscopic constant for water, \(K_f\), is 1.86 °C/molal.

Rearrange the freezing-point depression equation to solve for the molality of the solute (ethylene glycol): \[ m = \frac{\Delta T_\text{f}}{K_\text{f} \cdot i} \] Next, plug the given values into the equation: \(ΔT_\text{f} = - (-5.00 °\text{C}) = 5.00 °\text{C}\) (since we are looking for a depression of 5 degrees) \(K_\text{f} = 1.86 °\text{C/molal}\) \(i = 1\) \[ m = \frac{5.00 °\text{C}}{1.86 °\text{C/molal} \cdot 1} = 2.69 \,\text{molal} \] So the molality of ethylene glycol required to achieve the desired freezing point is 2.69 molal.

Now, we need to find the mass of ethylene glycol in grams. Use the formula \[ \text{mass} = \text{molality} \times \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] The molar mass of ethylene glycol (\(\text{C}_2 \text{H}_6\text{O}_2\)) is: \[ (2 \times 12.01\,\text{g/mol}) + (6 \times 1.01\,\text{g/mol}) + (2 \times 16.00\,\text{g/mol}) = 62.07\,\text{g/mol} \] Now, plug in the values we found: \(m = 2.69 \, \text{molal}\) \(M_\text{ethylene glycol} = 62.07 \,\text{g/mol}\) \(\text{mass of water} = 1000\, \text{g}=1.00 \,\text{kg}\) \[ \text{mass} = 2.69 \,\text{molal} \times \frac{62.07 \,\text{g/mol}}{1.00 \,\text{kg}} \] \[ \text{mass} = 167 \,\text{g} \] So, approximately 167 grams of ethylene glycol must be added to 1.00 kg of water to produce a solution that freezes at -5.00 °C.