A glucose solution contains 55.8g of glucose (C6H12O6) in 455g of water. Calculate the freezing poing and boiling point of the solution. (Assume a density of 1.00 g/ml for water.
This what I did, I got moles of C6H12O6, 0.89 moles C6H12O6
then I got the mass of the solution 455g/1000kg, 0.455 I then divided 0.89 solute mol/0.455 kg solvent. I was going to do the formula for freezing and boiling but what do I do with the density. I hope I havent confused you. I guess I don't know what to do with the density or if I have set this up right.
You ignore the density. Since it is 1.00 g/mL, that just means that you could calculate the volume of the solution but you don't need that.
moles glucose = 55.8/180.16 = not 0.89 but closer to 0.3 but that isn't exact.
m = moles/kg solvent = 0.3/0.455 = x
delta T = Kf*m and
delta T = Kb*m
Well, it seems like you're on the right track! To calculate the freezing and boiling points of the solution, you need to use the formula:
ΔT = K * m
Where ΔT is the change in temperature, K is the molal freezing/boiling point constant, and m is the molality of the solution (moles of solute per kilogram of solvent).
To calculate the molality, you correctly found the moles of glucose (C6H12O6) in the solution, which is 0.89 moles. And you correctly converted the mass of water to the mass of the solvent, which is 0.455 kg.
Now, to find the molality of the solution, m, you simply divide the moles of glucose by the mass of the water:
m = 0.89 mol / 0.455 kg = 1.956 mol/kg
Since you know the molality, you can now calculate the freezing and boiling point changes using the freezing/boiling point constants.
For the freezing point, use the formula:
ΔT(freezing) = K(freezing) * m
And for the boiling point, use the formula:
ΔT(boiling) = K(boiling) * m
The specific values for the freezing and boiling point constants, K, depend on the solvent used. For water, K(freezing) is approximately -1.86 °C/m and K(boiling) is approximately 0.51 °C/m.
So, plug in the values and solve for ΔT(freezing) and ΔT(boiling) to find the change in temperature. Then, you can add or subtract this change from the normal freezing and boiling points of water (0 °C and 100 °C, respectively) to get the freezing and boiling points of the solution.
Hope this helps you get through this icy and steamy situation!
To calculate the freezing point and boiling point of the solution, you need to use the molality of the solution, which is defined as moles of solute per kilogram of solvent.
First, you correctly calculated the moles of glucose (C6H12O6) in the solution as 0.89 moles.
To calculate the mass of the solvent (water), you divided 455g by 1000 to convert it to 0.455 kg.
Now, you need to calculate the molality of the solution, which is the ratio of moles of solute to the mass of the solvent in kg.
Molality = moles of solute / mass of solvent in kg
Molality = 0.89 / 0.455
Molality = 1.95 mol/kg
Now, you can use the molality to calculate the freezing point and boiling point of the solution.
Freezing point depression (ΔTf) and boiling point elevation (ΔTb) are given by the following equations:
ΔTf = Kf * molality
ΔTb = Kb * molality
Here, Kf and Kb are the freezing point depression constant and boiling point elevation constant, respectively. For water, Kf = 1.86 °C/m and Kb = 0.512 °C/m.
Calculating ΔTf:
ΔTf = 1.86 * 1.95
ΔTf ≈ 3.627 °C
Calculating ΔTb:
ΔTb = 0.512 * 1.95
ΔTb ≈ 0.9996 °C
To find the freezing point of the solution, subtract ΔTf from the freezing point of pure water (0 °C):
Freezing point = 0 °C - 3.627 °C
Freezing point ≈ -3.627 °C
To find the boiling point of the solution, add ΔTb to the boiling point of pure water (100 °C):
Boiling point = 100 °C + 0.9996 °C
Boiling point ≈ 100.9996 °C
Therefore, the freezing point of the solution is approximately -3.627 °C and the boiling point is approximately 100.9996 °C.
To calculate the freezing point and boiling point of the solution, you need to use the concept of colligative properties. One such property is the freezing point depression and boiling point elevation, which depend on the molality of the solution. Molality (m) is defined as moles of solute per kilogram of solvent.
First, let's calculate the molality (m) of the solution using the given information:
1. Calculate moles of glucose (C6H12O6):
Moles of glucose = mass of glucose / molar mass of glucose
Moles of glucose = 55.8 g / (6*12.01 g/mol + 12*1.01 g/mol + 6*16.00 g/mol)
Moles of glucose ≈ 0.93 mol
2. Calculate the mass of the solvent (water):
Mass of water = 455 g
3. Calculate molality:
Molality (m) = Moles of solute / Mass of solvent (in kg)
Molality (m) = 0.93 mol / 0.455 kg
Molality (m) ≈ 2.04 mol/kg
Now, let's calculate the freezing point depression and boiling point elevation using the following formulas:
Freezing Point Depression:
ΔTf = -Kf * m * i
Boiling Point Elevation:
ΔTb = Kb * m * i
Where:
ΔTf = Change in freezing point
ΔTb = Change in boiling point
Kf = Freezing point constant (for water, Kf = 1.86 °C/m)
Kb = Boiling point constant (for water, Kb = 0.512 °C/m)
m = Molality of the solution
i = Van't Hoff factor (the number of particles the solute breaks into in the solution; for glucose, i = 1)
Calculating the freezing point depression:
ΔTf = -1.86 °C/m * 2.04 mol/kg * 1
ΔTf ≈ -3.80 °C
Calculating the boiling point elevation:
ΔTb = 0.512 °C/m * 2.04 mol/kg * 1
ΔTb ≈ 1.04 °C
To get the actual freezing point and boiling point of the solution, you need to add/subtract the calculated values from the normal freezing point (0 °C) and boiling point (100 °C) of pure water, respectively.
Freezing point of the solution:
Freezing point = Normal freezing point - ΔTf
Freezing point ≈ 0 °C - (-3.80 °C)
Freezing point ≈ 3.80 °C
Boiling point of the solution:
Boiling point = Normal boiling point + ΔTb
Boiling point ≈ 100 °C + 1.04 °C
Boiling point ≈ 101.04 °C
Therefore, the freezing point of the solution is approximately 3.80 °C, and the boiling point of the solution is approximately 101.04 °C.