what is the freezing point depression when 153g of bromide is added to 1000g of benzene?
Bromide or Bromine? Which bromide?
what is the freezing point depression when 153g of bromine is added to 1000g of benzene?
To calculate the freezing point depression, we need to know the molal freezing point depression constant of the solvent and the molality of the solute.
The freezing point depression can be determined using the following equation:
∆T = Kf * m
Where:
∆T is the freezing point depression,
Kf is the molal freezing point depression constant of the solvent,
and m is the molality of the solute.
First, we need to find the molality of the solute, bromide (Br-), which is the moles of solute dissolved per kilogram of solvent.
Step 1: Convert the given mass of bromide (153g) into moles.
To do this, we need to divide the mass by the molar mass of bromide (79.9 g/mol).
Number of moles = mass / molar mass
Number of moles = 153g / 79.9 g/mol
Let's calculate the number of moles of bromide:
Number of moles = 1.91 moles
Step 2: Calculate the molality of the solute.
The molality (m) is given by the formula:
Molality = moles of solute / mass of solvent (in kg)
In this case, the mass of the solvent, benzene, is given as 1000g. We need to convert it to kilograms:
Mass of solvent = 1000g / 1000 = 1 kg
Now we can calculate the molality:
Molality = 1.91 moles / 1 kg
Molality = 1.91 mol/kg
Step 3: Determine the molal freezing point depression constant (Kf) of benzene.
The molal freezing point depression constant for benzene is approximately 5.12 °C⋅kg/mol.
Step 4: Calculate the freezing point depression (∆T).
Using the values obtained in the previous steps:
∆T = Kf * m
∆T = 5.12 °C⋅kg/mol * 1.91 mol/kg
Let's calculate the freezing point depression:
∆T = 9.7932 °C
Therefore, the freezing point depression when 153g of bromide is added to 1000g of benzene is approximately 9.7932 °C.