James is cooking pasta and adds 1.40 g of NaCl (NaCl = 58.5 g/mol) to 75.6 g of water, in which the NaCl dissociates completely. Calculate the solution's new freezing and boiling point if the kf for water is 1.86 0C/m and its kb is 0.512 0C/m.
Freezing point = -1.178°C, boiling point = 100.3°C
Freezing point = -0.00118 °C, boiling point = 100.1°C
Freezing point = -0.325°C, boiling point = 101.2°C
Freezing point = -0.589°C, boiling point = 103.2°C
Freezing point = 5.000°C, boiling point = 105.1 °C
To solve this problem, we need to use the formulas of colligative properties: ΔTf = Kf * m and ΔTb = Kb * m.
First, we need to calculate the number of moles of NaCl:
Number of moles of NaCl = mass of NaCl / molar mass of NaCl
Number of moles of NaCl = 1.40 g / 58.5 g/mol
Number of moles of NaCl = 0.0239 mol
Next, we need to calculate the molality (moles of solute per kilogram of solvent):
Molality = number of moles of NaCl / mass of water in kg
Mass of water in kg = 75.6 g / 1000 g/kg
Mass of water in kg = 0.0756 kg
Molality = 0.0239 mol / 0.0756 kg
Molality = 0.316 mol/kg
Now we can use the formulas for ΔTf and ΔTb to calculate the changes in freezing and boiling points:
ΔTf = Kf * m
ΔTf = 1.86 °C/m * 0.316 mol/kg
ΔTf = 0.58776 °C
ΔTb = Kb * m
ΔTb = 0.512 °C/m * 0.316 mol/kg
ΔTb = 0.16192 °C
Finally, we can calculate the new freezing and boiling points:
Freezing point = original freezing point - ΔTf
Freezing point = 0°C - 0.58776 °C
Freezing point = -0.58776 °C
Boiling point = original boiling point + ΔTb
Boiling point = 100°C + 0.16192°C
Boiling point = 100.16192 °C
Therefore, the solution's new freezing point is -0.58776 °C and the boiling point is 100.16192 °C.