what is the freezing point of a solution of 12.0 g of CCl4 dissolved in 750.0g of benzene? the freezing point of benzene is 5.48 degrees C ; Kf is 5.12 degrees c/m.
Well, you've certainly found yourself in a chilly situation! To determine the freezing point of this solution, we can use the formula ΔT = Kf * m, where ΔT is the change in temperature, Kf is the freezing point depression constant, and m is the molality of the solution.
First, let's calculate the molality (m) of the solution. We know that the molality is determined by dividing the moles of the solute by the mass of the solvent in kilograms. Since we're dealing with grams, we should convert the mass of benzene to kilograms.
Mass of benzene = 750.0g = 0.75kg
Now, let's calculate the moles of CCl4. We can do this by dividing the mass of CCl4 by its molar mass.
Molar mass of CCl4 = 12.0g/mol + (4 * 35.453g/mol) = 153.823g/mol
Moles of CCl4 = 12.0g / 153.823g/mol = 0.0779 mol
Now we can calculate the molality:
m = moles of solute / mass of solvent in kg
m = 0.0779 mol / 0.75 kg = 0.104 M
Now let's calculate the freezing point depression:
ΔT = Kf * m
ΔT = 5.12°C/m * 0.104 M = 0.5328°C
Finally, to find the freezing point of the solution, we subtract the freezing point depression from the freezing point of benzene:
Freezing point of solution = Freezing point of benzene - ΔT
Freezing point of solution = 5.48°C - 0.5328°C ≈ 4.95°C
So the freezing point of the solution of 12.0 g of CCl4 dissolved in 750.0 g of benzene is approximately 4.95°C. Stay frosty!
To find the freezing point of a solution, we first need to calculate the molality (m) of the solution using the formula:
molality (m) = moles of solute / mass of solvent (in kg)
Step 1: Convert the mass of solute (CCl4) from grams to moles.
To do this, we need to know the molar mass of CCl4, which is:
CCl4 = (1 x 12.01 g/mol) + (4 x 35.45 g/mol) = 153.83 g/mol
Using the molar mass, we can calculate the number of moles of CCl4:
moles of CCl4 = mass of CCl4 / molar mass of CCl4
moles of CCl4 = 12.0 g / 153.83 g/mol
Step 2: Convert the mass of the solvent (benzene) from grams to kilograms.
mass of benzene = 750.0 g / 1000 = 0.750 kg
Step 3: Calculate the molality of the solution.
molality (m) = moles of solute / mass of solvent (in kg)
molality (m) = moles of CCl4 / mass of benzene
Step 4: Calculate the change in freezing point (∆T)
∆T = Kf * m
Step 5: Calculate the freezing point of the solution.
freezing point of the solution = freezing point of the solvent - ∆T
Now, let's plug in the values:
moles of CCl4 = 12.0 g / 153.83 g/mol
mass of benzene = 0.750 kg
Kf = 5.12 °C/m
molality (m) = (12.0 g / 153.83 g/mol) / 0.750 kg
∆T = 5.12 °C/m * molality (m)
freezing point of the solution = 5.48 °C - ∆T
By following these steps and plugging in the given values, you should be able to find the freezing point of the solution.
convert 12.0 g CCl4 to moles. moles = grams/molar mass
molality = moles/kg solvent
solve for molality
delta T = Kf*m.
Solve for delta T and subtract from the normal freezing point of benzene.
4.948
1. convert 12.0 g CCl4 to moles. moles = grams / molar mass :
12.0 g CCl4 / 153.812 (g/mol) = 0.078 mol CCl4
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2. find the concentration (expressed in molality) of the solution. molality (m) = moles solute / kg solvent :
m = 0.078 mol CCl4 / 750.0 kg C6H6 [AKA benzene]
m = 0.104
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3. find the freezing point depression (refers to ΔTf) of the solution when compared to the solvent. ΔTf = Kf • m :
ΔTf = 5.12 ºC/m • 0.014m
ΔTf = 0.532ºC
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4. FINALLY, find the freezing point of the solution. Tf (the freezing point) of the solution is equal to Tf (the freezing point) of the solvent [benzene AKA C6H6] minus ΔTf :
5.48ºC - 0.532ºC =
4.948ºC
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II FINAL SOLUTION