Calculate the freezing point and boiling point of a solution that contains 55.4 NaCl and 42.3g KBr dissolved in 750.3 ml H2O
I showed you how in the answer to your first post. All of these follow the same method.
To calculate the freezing point and boiling point of a solution, we need to use the concept of colligative properties. These properties depend on the number of solute particles present in the solution, not the identity of the solute itself.
First, we need to determine the number of moles of NaCl and KBr in the solution using their molar masses.
The molar mass of NaCl (sodium chloride) is 58.44 g/mol.
Moles of NaCl = mass of NaCl / molar mass of NaCl
Moles of NaCl = 55.4 g / 58.44 g/mol
Moles of NaCl = 0.949 mol
Similarly, the molar mass of KBr (potassium bromide) is 119 g/mol.
Moles of KBr = mass of KBr / molar mass of KBr
Moles of KBr = 42.3 g / 119 g/mol
Moles of KBr = 0.355 mol
Next, we need to convert the volume of water (750.3 ml) into liters (L) for our calculations.
Volume of water = 750.3 ml = 750.3 / 1000 = 0.7503 L
Now, we can calculate the total number of moles of solute in the solution by adding the moles of NaCl and KBr together.
Total moles of solute = Moles of NaCl + Moles of KBr
Total moles of solute = 0.949 mol + 0.355 mol
Total moles of solute = 1.304 mol
The next step is to calculate the molality (moles of solute per kilogram of solvent) of the solution. We need to know the mass of the water in kilograms.
Mass of water = volume of water (in liters) x density of water
Mass of water = 0.7503 L x 1 g/mL = 0.7503 kg
Molality (m) = moles of solute / mass of solvent (in kg)
Molality = 1.304 mol / 0.7503 kg
Molality = 1.738 mol/kg
Now, we can use the equation for the freezing point depression and boiling point elevation to calculate the changes in temperature.
Freezing point depression (∆Tf) = Kf x molality
Boiling point elevation (∆Tb) = Kb x molality
The freezing point depression constant (Kf) for water is 1.86 °C/m, and the boiling point elevation constant (Kb) for water is 0.512 °C/m.
∆Tf = 1.86 °C/m x 1.738 mol/kg
∆Tf = 3.23268 °C
∆Tb = 0.512 °C/m x 1.738 mol/kg
∆Tb = 0.889856 °C
To find the freezing point, subtract the freezing point depression (∆Tf) from the normal freezing point of water (0 °C).
Freezing point = 0 °C - 3.23268 °C
Freezing point = -3.23 °C
To find the boiling point, add the boiling point elevation (∆Tb) to the normal boiling point of water (100 °C).
Boiling point = 100 °C + 0.889856 °C
Boiling point = 100.89 °C
Therefore, the freezing point of the solution is approximately -3.23 °C, and the boiling point is approximately 100.89 °C.