For a 0.50 molal solution of sucrose in water, calculate the freezing point and the boiling point of the solution. Please list each step.?
dT = Kf*m or dT = Kb*m
You know m, look up Kf and Kb, solve for dT in each case.
For the new freezing point subtract from the normal freezing point.
For the new b.p., add to the normal boiling point. Post your work if you get stuck.
For a 0.50 molal solution of sucrose in water, calculate the freezing point and the boiling point of the solution
To calculate the freezing point and boiling point of a 0.50 molal solution of sucrose in water, you can follow these steps:
Step 1: Determine the molality of the solution.
Molality (m) is defined as moles of solute per kilogram of solvent. In this case, the solute is sucrose and the solvent is water. Given that the solution is 0.50 molal, it means there are 0.50 moles of sucrose dissolved in 1 kg of water.
Step 2: Calculate the molecular weight of sucrose.
The molecular weight of sucrose (C12H22O11) can be found using the periodic table. The atomic weights are:
C = 12.01 g/mol, H = 1.01 g/mol, and O = 16.00 g/mol.
Molecular weight of C12H22O11 = 12.01 * 12 + 1.01 * 22 + 16.00 * 11 = 342.34 g/mol.
Step 3: Calculate the mass of sucrose in the solution.
Since the molality is given as 0.50 molal, it means there are 0.50 moles of sucrose in 1 kg of water. To find the mass of sucrose, multiply the molecular weight by the number of moles:
Mass of sucrose = 0.50 mol * 342.34 g/mol = 171.17 g.
Step 4: Calculate the new freezing point of the solution.
The freezing point depression can be calculated using the equation: ΔTf = Kf * m, where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution.
For water, the cryoscopic constant is approximately 1.86 °C/m.
ΔTf = 1.86 °C/m * 0.50 mol/kg = 0.93 °C.
To calculate the new freezing point, subtract the freezing point depression from the normal freezing point of water, which is 0 °C:
Freezing point of the solution = 0 °C - 0.93 °C = -0.93 °C.
Therefore, the freezing point of the solution is -0.93 °C.
Step 5: Calculate the new boiling point of the solution.
The boiling point elevation can be calculated using the equation: ΔTb = Kb * m, where ΔTb is the change in boiling point, Kb is the ebullioscopic constant, and m is the molality of the solution.
For water, the ebullioscopic constant is approximately 0.512 °C/m.
ΔTb = 0.512 °C/m * 0.50 mol/kg = 0.256 °C.
To calculate the new boiling point, add the boiling point elevation to the normal boiling point of water, which is 100 °C:
Boiling point of the solution = 100 °C + 0.256 °C = 100.256 °C.
Therefore, the boiling point of the solution is 100.256 °C.
In summary:
- Freezing point of the solution: -0.93 °C.
- Boiling point of the solution: 100.256 °C.
To calculate the freezing point and boiling point of a 0.50 molal solution of sucrose in water, you need to follow a step-by-step process. Here are the steps:
Step 1: Determine the molality of the solution.
- Molality (m) is defined as the number of moles of solute per kilogram of solvent.
- In this case, the molality is given as 0.50 molal, which means there are 0.50 moles of sucrose per kilogram of water.
Step 2: Determine the freezing point depression constant (Kf) and boiling point elevation constant (Kb) for water.
- These constants depend on the solvent used.
- For water, the freezing point depression constant (Kf) is approximately 1.86 °C/m, and the boiling point elevation constant (Kb) is approximately 0.52 °C/m.
Step 3: Calculate the freezing point depression.
- The freezing point depression is given by the formula: ΔTf = Kf * m
- ΔTf represents the change in freezing point (compared to the pure solvent).
- Kf is the freezing point depression constant, and m is the molality of the solution.
- In this case, ΔTf = (1.86 °C/m) * (0.50 mol/kg) = 0.93 °C
Step 4: Calculate the freezing point of the solution.
- The freezing point of the solution can be obtained by subtracting the freezing point depression from the freezing point of the pure solvent.
- The freezing point of pure water is 0 °C, so the freezing point of the solution would be 0 °C - (0.93 °C) = -0.93 °C.
Step 5: Calculate the boiling point elevation.
- The boiling point elevation is given by the formula: ΔTb = Kb * m
- ΔTb represents the change in boiling point (compared to the pure solvent).
- Kb is the boiling point elevation constant, and m is the molality of the solution.
- In this case, ΔTb = (0.52 °C/m) * (0.50 mol/kg) = 0.26 °C
Step 6: Calculate the boiling point of the solution.
- The boiling point of the solution can be obtained by adding the boiling point elevation to the boiling point of the pure solvent.
- The boiling point of pure water is 100 °C, so the boiling point of the solution would be 100 °C + (0.26 °C) = 100.26 °C.
Therefore, for a 0.50 molal solution of sucrose in water, the freezing point would be approximately -0.93 °C and the boiling point would be approximately 100.26 °C.