what is the freezing point of 0.05 M glucose(C6H12O6), 0.05 M NaCl, and 0.05 M MgCl2?
To determine the freezing point of a solution, we can use the formula for freezing point depression:
∆Tf = Kf * m
Where:
∆Tf = freezing point depression
Kf = cryoscopic constant (which is specific for each solvent)
m = molality of the solution (moles of solute per kilogram of solvent)
For glucose(C6H12O6):
Molar mass of glucose = 180.156 g/mol
To find the molality, we need to find the moles of glucose and the mass of solvent (water) present.
Given:
Concentration of glucose = 0.05 M
To convert concentration to moles of solute per kg of solvent, we need to know the density of the solution or the mass of the solvent. Assuming we have 1 liter of solution, the mass of water (solvent) can be calculated.
Density of water = 1 g/mL
Mass of 1 liter of water = 1000 g
Since 1 liter of solution contains 0.05 moles of glucose, we can assume that it consists mostly of water. Therefore, we can consider the mass of the solvent to be 1000 g.
Molality of glucose (m) = moles of glucose / kg of solvent
= 0.05 moles / 1 kg
= 0.05 mol/kg
Now, we need to find the freezing point depression (∆Tf) for glucose.
Next, we need the cryoscopic constant (Kf) for water. For water, the cryoscopic constant is approximately 1.86 °C/m.
Finally, we can use the formula to calculate the freezing point depression (∆Tf).
∆Tf (glucose) = Kf * m (glucose)
= 1.86 °C/m * 0.05 mol/kg
= 0.093 °C
So, the freezing point of a 0.05 M glucose (C6H12O6) solution is reduced by approximately 0.093 °C.
To find the freezing point depression for the sodium chloride (NaCl) and magnesium chloride (MgCl2) solutions, we follow the same procedure and use the relevant cryoscopic constants.
Cryoscopic constant (Kf) for NaCl in water is approximately 1.86 °C/m.
Cryoscopic constant (Kf) for MgCl2 in water is approximately 3.61 °C/m.
Assuming the molar mass of NaCl as 58.44 g/mol and MgCl2 as 95.211 g/mol, we can calculate the molality (m) for the NaCl and MgCl2 solutions.
Molality of NaCl (m) = 0.05 moles / 1 kg
= 0.05 mol/kg
Molality of MgCl2 (m) = 0.05 moles / 1 kg
= 0.05 mol/kg
Now, we can calculate the freezing point depression (∆Tf) for NaCl and MgCl2 using the formula:
∆Tf (NaCl) = Kf * m (NaCl)
= 1.86 °C/m * 0.05 mol/kg
= 0.093 °C
∆Tf (MgCl2) = Kf * m (MgCl2)
= 3.61 °C/m * 0.05 mol/kg
= 0.181 °C
So, the freezing point of a 0.05 M NaCl solution is reduced by approximately 0.093 °C, and the freezing point of a 0.05 M MgCl2 solution is reduced by approximately 0.181 °C.
To determine the freezing point of a solution, we need to utilize the concept of colligative properties. These properties depend on the concentration of solute particles, regardless of their identity.
The freezing point depression (∆Tf) formula for a solution can be given as:
∆Tf = Kf * m
Where:
∆Tf = Freezing point depression
Kf = Freezing point depression constant (molality constant)
m = Molality of the solution (moles of solute per kg of solvent)
First, we need to calculate the molality for each solution:
Molality (m) = moles of solute / mass of solvent (in kg)
1. Glucose (C6H12O6):
Molar mass of glucose (C6H12O6) = 180.18 g/mol
Moles of glucose = 0.05 M * (180.18 g/mol) * 1 L = 0.05 mol
Assuming the total volume of the solution is 1 liter, the mass of the solvent (water) can be considered as 1000 g or 1 kg. Therefore:
Molality (m) = 0.05 mol / 1 kg = 0.05 mol/kg
2. NaCl:
Molar mass of NaCl = 58.44 g/mol
Moles of NaCl = 0.05 M * (58.44 g/mol) * 1 L = 0.05 mol
Molality (m) = 0.05 mol / 1 kg = 0.05 mol/kg
3. MgCl2:
Molar mass of MgCl2 = (24.31 g/mol) + 2 * (35.45 g/mol) = 95.21 g/mol
Moles of MgCl2 = 0.05 M * (95.21 g/mol) * 1 L = 0.05 mol
Molality (m) = 0.05 mol / 1 kg = 0.05 mol/kg
Now that we have the molality values for each solution, we need to know the freezing point depression constant (Kf) for water. For water, the experimental value of Kf is approximately 1.86 °C·kg/mol.
Using the formula mentioned earlier, we can calculate the freezing point depression (∆Tf) for each solution:
1. Glucose (C6H12O6):
∆Tf (Glucose) = 1.86 °C·kg/mol * 0.05 mol/kg = 0.093 °C
2. NaCl:
∆Tf (NaCl) = 1.86 °C·kg/mol * 0.05 mol/kg = 0.093 °C
3. MgCl2:
∆Tf (MgCl2) = 1.86 °C·kg/mol * 0.05 mol/kg = 0.093 °C
Therefore, the freezing point of 0.05 M glucose (C6H12O6), 0.05 M NaCl, and 0.05 M MgCl2 will be depressed by approximately 0.093 °C.
Here is a good lesson in chemistry and you need to learn it and put it practice today. Find the caps key and know when to use it. You certainly need it to start sentences as in the What above. You do NOT need it when you intend to write molality. The small m stands for molality; the cap M stands for molarity. So I assume you mean 0.05 m glucose, 0.05 m NaCl, and 0.05 m MgCl2.
One other point, you're looking for the freezing point of what? water? benzene? napthalene? just what?
delta T = i*Kf*m
Substitute and solve for delta T, then subtract that from the normal freezing point of the solvent.
i = van't Hoff factor.
i = 1 for glucose
i = 2 for NaCl
i = 3 for MgCl2