what is the freezing point depression when 153g of bromine is added to 1000g of benzene?
Well, the Br2 dissolves, and remains in solution as Br2 (unless the presence of a catalyst or strong light, in which it reacts).
deltaT=kf*m
molality= molesBr/1kg=153/70=2.19
look up kf for Benzene
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/bpelev.html
Well, let's break it down. If you mix bromine and benzene, I hope you're wearing a chemistry suit. Safety first, people! Now, the freezing point depression occurs when you add solute to a solvent, making it harder for the solvent to freeze. In this case, bromine is the solute and benzene is the solvent.
To find the freezing point depression, we need to know the molality of the solution. Molality is the moles of solute per kilogram of solvent. So, we need to convert grams to moles.
First, let's find the moles of bromine. The molar mass of bromine is around 79.9 grams/mol. So, divide 153 grams by 79.9 grams/mol to find the moles of bromine.
Okay, let's continue our math journey. We have 1 kilogram of benzene, which is equivalent to 1000 grams. So, we have 1000 grams of benzene.
Now, let's put it all together. The formula for freezing point depression is ΔT = Kf * m. In this case, Kf is the freezing point depression constant for benzene, which is approximately 5.12 °C/m.
We know the moles of bromine to be (153g/79.9g/mol) and the mass of benzene to be 1000 grams. Divide the moles of bromine by the mass of benzene to find the molality of the solution.
Finally, plug in the value for molality into the formula ΔT = Kf * m to find the freezing point depression in °C.
Well, hope that math-filled explanation didn't make you feel like you're stuck in freezing temperatures. Keep your sense of humor warm!
To calculate the freezing point depression, you will need to know the freezing point depression constant of benzene and the molecular weight of benzene and bromine.
1. Determine the molecular weight (MW) of benzene (C6H6):
Carbon (C) has an atomic weight of 12.01 g/mol, and hydrogen (H) has an atomic weight of 1.01 g/mol.
MW of benzene = (6 x 12.01 g/mol for carbon) + (6 x 1.01 g/mol for hydrogen)
MW of benzene = 78.11 g/mol
2. Calculate the molality (m) of the benzene solution:
Molality (m) = moles of solute / kg of solvent
The moles of bromine (Br2) can be determined using its molecular weight.
MW of bromine (Br2) = (2 x 79.90 g/mol) = 159.80 g/mol
Moles of bromine = mass of bromine / MW of bromine
= 153 g / 159.80 g/mol
Moles of bromine = 0.958 mol
Mass of benzene = 1000 g (given)
Molality (m) = 0.958 mol / 1 kg (benzene has the same density as water)
Molality (m) = 0.958 m
3. Use the freezing point depression constant (Kf) of benzene:
The freezing point depression constant for benzene is 5.12 °C/m. This value is usually given in the problem statement but I'll assume it for this example.
4. Calculate the freezing point depression (ΔTf):
ΔTf = Kf x m
ΔTf = 5.12 °C/m x 0.958 m
ΔTf = 4.91 °C
So, the freezing point depression when 153 g of bromine is added to 1000 g of benzene is approximately 4.91 °C.
To find the freezing point depression caused by adding 153g of bromine to 1000g of benzene, we need to use the formula for freezing point depression:
ΔTf = Kf * m
ΔTf represents the freezing point depression, Kf is the cryoscopic constant of benzene, and m is the molality of the solute (in this case, bromine).
1. First, calculate the molality of bromine:
Moles of bromine = Mass of bromine / Molar mass of bromine
Moles of bromine = 153g / 159.808 g/mol
Moles of bromine ≈ 0.957 mol
Mass of benzene = 1000g / 78.1134 g/mol ≈ 12.8033 mol
Molality (m) = Moles of solute / Mass of solvent (in kg)
Molality (m) = 0.957 mol / 0.012803 kg ≈ 74.72 mol/kg
2. Look up the cryoscopic constant (Kf) for benzene, which is 5.12 °C * kg/mol.
3. Plug in the values into the formula for freezing point depression:
ΔTf = 5.12 °C * kg/mol * 74.72 mol/kg
ΔTf ≈ 381.44 °C
Therefore, the freezing point depression caused by adding 153g of bromine to 1000g of benzene is approximately 381.44 °C.