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.
To calculate the new freezing and boiling points of the solution, we need to use the equation:
ΔT = i * Kf * m
where:
ΔT is the change in temperature (in degrees Celsius)
i is the van't Hoff factor
Kf is the freezing point depression constant for the solvent (in degrees Celsius/mole)
m is the molality of the solute (in moles of solute per kg of solvent)
First, we need to calculate the molality (m) of the NaCl solution.
The molar mass of NaCl is 58.5 g/mol, so the number of moles of NaCl is:
moles of NaCl = mass of NaCl / molar mass of NaCl
moles of NaCl = 1.40 g / 58.5 g/mol = 0.024 mol
The mass of water (solvent) is given as 75.6 g.
Converting the mass of water to kg:
mass of water = 75.6 g / 1000 = 0.0756 kg
Next, we need to calculate the molality of the NaCl solution:
molality (m) = moles of solute / mass of solvent (in kg)
molality (m) = 0.024 mol / 0.0756 kg = 0.317 mol/kg
Now we can calculate the freezing point depression (ΔTf):
ΔTf = i * Kf * m
For NaCl, the van't Hoff factor (i) is equal to 2 because NaCl dissociates completely in water to form 2 ions (Na+ and Cl-).
ΔTf = 2 * 1.86 0C/m * 0.317 mol/kg
ΔTf = 1.16 0C
The freezing point of pure water is 0 0C, so the new freezing point of the NaCl solution is:
Freezing point = 0 0C - 1.16 0C = -1.16 0C
Next, we can calculate the boiling point elevation (ΔTb):
ΔTb = i * Kb * m
For NaCl, the van't Hoff factor (i) is equal to 2 because NaCl dissociates completely in water to form 2 ions (Na+ and Cl-).
ΔTb = 2 * 0.512 0C/m * 0.317 mol/kg
ΔTb = 0.325 0C
The boiling point of pure water is 100 0C, so the new boiling point of the NaCl solution is:
Boiling point = 100 0C + 0.325 0C = 100.325 0C
Therefore, the new freezing point of the solution is -1.16 0C and the new boiling point of the solution is 100.325 0C.