An aqueous solution is 0.257 m NaOH. What are the boiling point and freezing point of this solution?
delta Tf = i*Kf*m for freezing point.
delta Tb = i*Kb*m for boiling point.
For f.p., subtract delta T from normal freezing point.
For b.p., add delta T to normal boiling point.
To determine the boiling point and freezing point of a solution, we need to use the concept of colligative properties. The boiling point elevation and freezing point depression are two such properties that depend on the concentration of solute particles in a solution.
First, we need to calculate the molality (concentration) of the NaOH solution.
Molality (m) is defined as the moles of solute per kilogram of solvent. In this case, the solvent is water (H2O).
Given:
- The molarity (M) of the NaOH solution is 0.257 mol/L
- The molar mass (Mm) of NaOH is 39.997 g/mol
To calculate the molality, we need to convert the molarity to moles of solute.
Step 1: Calculate the moles of NaOH
moles of NaOH = molarity × volume of solution (in liters)
moles of NaOH = 0.257 mol/L × 1 L
moles of NaOH = 0.257 mol
Step 2: Calculate the mass of water (solvent)
Given that the solution is aqueous, we assume that the total volume is 1 L. Since the density of water is approximately 1 g/mL, the mass of water is 1000 g.
Step 3: Calculate the molality (m)
molality (m) = moles of solute / mass of solvent (in kg)
molality (m) = 0.257 mol / 1 kg = 0.257 mol/kg
Now, we can use the molality value to calculate the boiling point elevation and freezing point depression:
Boiling Point Elevation:
The boiling point elevation (ΔTb) is given by the equation:
ΔTb = Kb × m
where Kb is the molal boiling point constant (a constant for a given solvent) and m is the molality.
For water, the molal boiling point constant (Kb) is approximately 0.512 °C/m.
Substituting the values:
ΔTb = 0.512 °C/m × 0.257 mol/kg
ΔTb = 0.13184 °C
The boiling point of water at standard pressure is 100 °C. Hence, the boiling point of the NaOH solution is:
Boiling point = 100 °C + ΔTb
Boiling point = 100 °C + 0.13184 °C
Boiling point = 100.13184 °C
Freezing Point Depression:
The freezing point depression (ΔTf) is given by the equation:
ΔTf = Kf × m
where Kf is the molal freezing point constant (a constant for a given solvent) and m is the molality.
For water, the molal freezing point constant (Kf) is approximately 1.86 °C/m.
Substituting the values:
ΔTf = 1.86 °C/m × 0.257 mol/kg
ΔTf = 0.47742 °C
The freezing point of pure water at standard pressure is 0 °C. Hence, the freezing point of the NaOH solution is:
Freezing point = 0 °C - ΔTf
Freezing point = 0 °C - 0.47742 °C
Freezing point = -0.47742 °C
Therefore, the boiling point of the 0.257 m NaOH solution is approximately 100.13184 °C and the freezing point is approximately -0.47742 °C.