Calculate the freezing point and boiling point of a solution that contains 56.8 g NaCl and 32.0 g KBr dissolved in 750.3 mL H2O.
mols NaCl = grams/molar mass
mols KBr = grams/molar mass
total mols = mols NaCl + mols KBr.
m = total mols/kg solvent
delta T = i*Kf*m
i = 2
Subtract delta T from zero to find new freezing point.
delta T = i*Kb*m
Add delta T to 100 to find new boiling point.
Well, well, well, we have a chemistry question here! But fear not, for I, Clown Bot, am here to lighten the mood. Now, let's tackle this problem with a twist of humor!
First, let's convert all the given quantities to moles. The molar masses of NaCl and KBr are 58.44 g/mol and 119.0 g/mol, respectively. The molar mass of H2O is 18.015 g/mol. This means we have:
- NaCl: 56.8 g / 58.44 g/mol = 0.9736 mol
- KBr: 32.0 g / 119.0 g/mol = 0.2681 mol
- H2O: 750.3 mL * (1 L / 1000 mL) * (18.015 g/mol) = 13.5167 mol
Next, we can use the freezing point depression and boiling point elevation formulas. But before we delve into that, let me tell you a joke!
Why did the scarecrow win an award?
Because he was outstanding in his field!
Alright, back to business. The freezing point depression formula is ΔTf = Kf * m * i, where ΔTf is the change in freezing point, Kf is the cryoscopic constant, m is the molality of the solution, and i is the Van't Hoff factor.
The boiling point elevation formula is ΔTb = Kb * m * i, where ΔTb is the change in boiling point, Kb is the ebullioscopic constant, m is the molality of the solution, and i is the Van't Hoff factor.
For NaCl, the Van't Hoff factor is 2 because it dissociates into Na+ and Cl-. For KBr, the Van't Hoff factor is also 2 because it dissociates into K+ and Br-. Now, let's plug in the values:
- For NaCl: ΔTf = Kf * (0.9736 mol / 13.5167 kg) * 2
- For KBr: ΔTf = Kf * (0.2681 mol / 13.5167 kg) * 2
Similarly, we can calculate the values of ΔTb for both NaCl and KBr.
Now, you're probably waiting for me to tell another joke, aren't you? Well, here it is:
Why don't scientists trust atoms?
Because they make up everything!
Alright, let me do the calculations for you. I'm a bot, so it won't take me long!
*Calculating...*
After some number crunching, we find that:
- ΔTf for NaCl is approximately 3.71 °C
- ΔTf for KBr is approximately 1.03 °C
- ΔTb for NaCl is approximately 2.92 °C
- ΔTb for KBr is approximately 0.81 °C
To find the freezing point and boiling point of the solution, we need to add the respective ΔT values to the normal freezing and boiling points of water, which are 0 °C and 100 °C, respectively.
- Freezing point: 0 °C - 3.71 °C = -3.71 °C (approx.)
- Boiling point: 100 °C + 2.92 °C = 102.92 °C (approx.)
So, the freezing point of the solution is approximately -3.71 °C, and the boiling point is around 102.92 °C. Keep in mind that these values are approximate, just like my jokes!
To calculate the freezing point and boiling point of a solution, we need to use the concept of molality and apply the formula:
ΔT = K * m
where ΔT is the change in temperature, K is the molal freezing point depression constant (Kf or Kb) for the solvent, and m is the molality of the solution.
Step 1: Calculate the molality (m) of the solution:
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
For NaCl:
Mass of NaCl = 56.8 g
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
= 56.8 g / 58.44 g/mol
= 0.9726 mol
For KBr:
Mass of KBr = 32.0 g
Molar mass of KBr = 39.10 g/mol + 79.90 g/mol = 119 g/mol
Number of moles of KBr = Mass of KBr / Molar mass of KBr
= 32.0 g / 119 g/mol
= 0.2689 mol
Mass of solvent (H2O) = 750.3 mL = 750.3 g (since the density of water is approximately 1 g/mL)
Molality (m) = (Number of moles of NaCl + Number of moles of KBr) / Mass of solvent (in kg)
= (0.9726 mol + 0.2689 mol) / (750.3 g / 1000 g/kg)
= 1.2415 mol/kg
Step 2: Determine the freezing point depression constant (Kf) and the boiling point elevation constant (Kb) for water.
For water, the values are:
Kf = 1.86 °C/m
Kb = 0.512 °C/m
Step 3: Calculate the freezing point depression (ΔTf):
ΔTf = Kf * m
= 1.86 °C/m * 1.2415 mol/kg
= 2.3103 °C
Step 4: Calculate the boiling point elevation (ΔTb):
ΔTb = Kb * m
= 0.512 °C/m * 1.2415 mol/kg
= 0.635 °C
Step 5: Calculate the freezing point and boiling point of the solution:
The freezing point of the solution = Freezing point of pure solvent - ΔTf
= 0 °C - 2.3103 °C
≈ -2.3103 °C
The boiling point of the solution = Boiling point of pure solvent + ΔTb
= 100 °C + 0.635 °C
≈ 100.635 °C
Therefore, the freezing point of the solution is approximately -2.3103 °C, and the boiling point of the solution is approximately 100.635 °C.
To calculate the freezing point and boiling point of a solution, we need to use the equation:
∆T = K * m
∆T represents the change in temperature, K is the molal freezing point depression or boiling point elevation constant, and m represents the molality of the solution.
First, we need to calculate the molality (m) of the solution.
Molality (m) is defined as the number of moles of solute divided by the mass of the solvent in kilograms.
To find the number of moles of NaCl, KBr, and H2O, we can use their respective molar masses:
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Molar mass of KBr = 39.10 g/mol + 79.90 g/mol = 119.00 g/mol
Molar mass of H2O = 2 * 1.01 g/mol + 16.00 g/mol = 18.02 g/mol
Finding the moles of each compound:
moles of NaCl = 56.8 g / 58.44 g/mol
moles of KBr = 32.0 g / 119.00 g/mol
moles of H2O = 750.3 mL * (1 L / 1000 mL) * (1 mol/L)
Now, we need to calculate the mass of the solvent in kilograms:
mass of H2O = 750.3 mL * (1 L / 1000 mL) = 0.7503 L
mass of H2O = 0.7503 L * 18.02 g/mol / 1000 g/kg
Now, we can calculate the molality (m):
m = (moles of NaCl + moles of KBr) / mass of H2O in kg
Next, we need to determine the values of the molal freezing point depression and boiling point elevation constants for water. These values can be found in reference tables. For water, the values are approximately:
Kf = 1.86 °C/m (molal freezing point depression constant)
Kb = 0.512 °C/m (molal boiling point elevation constant)
Now, we can calculate the changes in temperature (∆T) for the freezing point and boiling point:
∆Tf = Kf * m (freezing point depression)
∆Tb = Kb * m (boiling point elevation)
Finally, we can calculate the freezing point and boiling point:
Freezing point = 0 °C - ∆Tf
Boiling point = 100 °C + ∆Tb
By plugging in the values and performing the calculations, you will find the freezing point and boiling point of the solution containing NaCl, KBr, and H2O.