a solution consisting of 5.0 gm of an organic solute per 25 gm of CCl4 boils at 81.5 degree celecius .what is the molecular weight of the solute? If the boiling point is 76.8 degree celecius and Kb of CCl4 is 5.02 .
delta T = Kb*m
Calculate m
m = moles/kg solvent.
Calculate moles
moles = grams/molar mass
calculate molar mass.
Note the correct spelling of celsius.
To find the molecular weight of the solute, we can use the formula:
∆Tb = Kbm
Where:
∆Tb is the boiling point elevation (the difference between the boiling point with solute and the boiling point without solute).
Kb is the boiling point elevation constant for the solvent (CCl4 in this case).
m is the molality of the solution, which is given by:
m = moles of solute / mass of solvent (in kg)
First, let's calculate the molality (m). Given that 5.0 g of solute is dissolved in 25 g of CCl4, we need to convert the mass of solvent to kg:
mass of solvent = 25 g = 25/1000 kg = 0.025 kg
Next, we calculate the moles of solute:
moles of solute = mass of solute / molecular weight of solute
We know that when 5.0 g of solute is dissolved in 0.025 kg of solvent, the molality is 5.0 g / 0.025 kg = 200 mol/kg.
Now we can rearrange the formula ∆Tb = Kbm and solve for the molecular weight (M) of the solute:
M = ∆Tb / (Kb x m)
Given that the boiling point elevation (∆Tb) is 81.5 - 76.8 = 4.7 °C, and the Kb of CCl4 is 5.02 °C/m, we can substitute these values into the formula:
M = 4.7 °C / (5.02 °C/m x 200 mol/kg)
Now we can calculate M:
M = 4.7 °C / (1004 °C·m·kg/mol)
M = 0.0047 mol·kg/mol
Therefore, the molecular weight of the solute is approximately 0.0047 g/mol.