Final answer:
The solution's freezing point will be -6.34°C, and the boiling point will be 101.74°C, accounting for the dissociation of NaCl into two ions.
Step-by-step explanation:
Molality (m) is defined as moles of solute per kilogram of solvent. There are 10 grams of NaCl, which is approximately 0.171 moles (10 g / 58.44 g/mol). As 100 ml of water has a mass of approximately 100 grams (or 0.1 kg), the molality of the solution is 1.71 mol/kg. Now, for NaCl, which dissociates into two ions, the van't Hoff factor (i) is 2. This means the effective molality is 3.42 mol/kg because you have twice as many particles affecting the properties of the solution.
The freezing point depression (ΔTf) is calculated using the formula ΔTf = i * Kf * m. Plugging in our values, we get ΔTf = 2 * 1.86°C/kg/mol * 1.71 mol/kg = 6.34°C. The normal freezing point of water is 0°C, so the solution's freezing point is 0°C - 6.34°C = -6.34°C.
For the boiling point elevation (ΔTb), we use the formula ΔTb = i * Kb * m. Therefore, ΔTb = 2 * 0.51°C/kg/mol * 1.71 mol/kg = 1.74°C. The normal boiling point of water is 100°C, hence the solution's boiling point is 100°C + 1.74°C = 101.74°C.