what are the normal freezing points and boiling points of the following solutions: a. 21.1g NaCl in 135mL of water b. 15.4g urea in 66.7 mL water?
Well, let me answer that with a bit of humor.
a. The normal freezing point of water is usually 0 degrees Celsius, but if you add 21.1 grams of NaCl, I'm afraid it becomes "salty cold." You might need some crackers to go with that freezing point.
b. As for the boiling point, water normally boils at 100 degrees Celsius, but if you add 15.4 grams of urea, it might turn into a "steamy surprise." Just make sure you don't forget to add a pinch of humor while cooking. Bon appé!
To calculate the freezing point and boiling point of a solution, we need to use the formula ΔT = Kf * m for freezing point depression and ΔT = Kb * m for boiling point elevation.
a. 21.1g NaCl in 135mL of water:
First, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute (NaCl) divided by the mass of the solvent (water) in kilograms.
Step 1: Calculate the moles of NaCl:
moles of NaCl = mass of NaCl / molar mass of NaCl
The molar mass of NaCl is 58.44 g/mol.
moles of NaCl = 21.1g NaCl / 58.44 g/mol
Step 2: Convert mL to kg:
mass of water = volume of water * density of water
The density of water is approximately 1g/mL.
mass of water = 135mL * 1g/mL
Now, we need to convert grams to kilograms:
mass of water = 135g / 1000g/kg
Step 3: Calculate molality:
molality (m) = moles of NaCl / mass of water (in kg)
Now, we can substitute the values into the formula to calculate the freezing point depression and boiling point elevation.
Step 4: Calculate the freezing point depression:
ΔT = Kf * m
The freezing point depression constant (Kf) for water is approximately 1.86 °C/m.
ΔT = 1.86 °C/m * molality (m)
Step 5: Calculate the boiling point elevation:
ΔT = Kb * m
The boiling point elevation constant (Kb) for water is approximately 0.512 °C/m.
ΔT = 0.512 °C/m * molality (m)
b. 15.4g urea in 66.7 mL water:
The process is similar to the previous example.
1. Calculate the moles of urea.
2. Convert mL to kg (for water).
3. Calculate molality (m) by dividing moles of urea by mass of water in kg.
4. Calculate the freezing point depression using the formula ΔT = Kf * m.
5. Calculate the boiling point elevation using the formula ΔT = Kb * m.
After completing these steps for both cases (a and b), you will have the freezing points and boiling points for the given solutions.
To find the normal freezing points and boiling points of the given solutions, we need to use the concept of colligative properties. Colligative properties depend on the concentration of solute particles in a solution rather than the nature of the solute itself.
The freezing point depression and boiling point elevation can be calculated using the following equations:
1. Freezing Point Depression:
ΔTf = Kf * molality
2. Boiling Point Elevation:
ΔTb = Kb * molality
In these equations:
- ΔTf represents the freezing point depression
- ΔTb represents the boiling point elevation
- Kf and Kb are the respective cryoscopic and ebullioscopic constants for the solvent
- molality is the concentration of the solute in mol/kg of solvent
Now let's calculate the molality for each solution:
a. 21.1g NaCl in 135mL of water:
- First, we need to convert the volume of water to kg by dividing by 1000: 135mL / 1000 = 0.135 kg
- Next, calculate the number of moles of NaCl: moles = mass / molar mass = 21.1g / (22.99g/mol + 35.45g/mol) ≈ 0.360 mol
- Finally, calculate the molality: molality = moles / kg of solvent = 0.360 mol / 0.135kg ≈ 2.67 mol/kg
b. 15.4g urea in 66.7 mL water:
- Convert the volume of water to kg: 66.7mL / 1000 = 0.0667 kg
- Calculate the number of moles of urea: moles = mass / molar mass = 15.4g / 60.06g/mol ≈ 0.256 mol
- Calculate the molality: molality = moles / kg of solvent = 0.256 mol / 0.0667 kg ≈ 3.84 mol/kg
Now that we have the molality values, we can use them to find the freezing and boiling point changes using the cryoscopic constant (Kf) and ebullioscopic constant (Kb) for water.
The cryoscopic constant (Kf) for water is approximately 1.86 °C/m, and the ebullioscopic constant (Kb) for water is approximately 0.512 °C/m.
a. Freezing Point Depression:
ΔTf = Kf * molality = 1.86 °C/m * 2.67 mol/kg ≈ 4.96 °C
b. Boiling Point Elevation:
ΔTb = Kb * molality = 0.512 °C/m * 3.84 mol/kg ≈ 1.97 °C
To find the normal freezing and boiling points of the solutions, we need to add or subtract the calculated temperature changes from the normal freezing point (-0.52 °C) and boiling point (100 °C) of pure water.
a. Normal Freezing Point:
-0.52 °C - 4.96 °C ≈ -5.48 °C
b. Normal Boiling Point:
100 °C + 1.97 °C ≈ 101.97 °C
Therefore, the normal freezing point of the 21.1g NaCl solution in 135mL water is approximately -5.48 °C, and the normal boiling point is approximately 101.97 °C.
For a.
moles NaCl = grams/molar mass = 21.1g/58.44 = ??
molality = moles/kg solvent = ??moles/0.135 g H2O = xx
Then delta T = i*Kf*m
i = 2; Kf = 1.86; m = xx
This will give you delta T and from that and knowing H2O usually freezes at 0C, calculate freezing point of the solution.
part b is done the same way; however, i = 1.