What is the boiling point of a solution of benzene contain 0.2 moles of a non volatile solute in 125 grams of benzene given that pure benzene has a b.p. of 80.1 Celcius and a Kbp of +2.53 C/m.
Thanks!
m = molality = mols/kg solvent. You know mols and kg solvent, solve for m
Then substitute m into the below equation.
delta T = Kb*m
Solve for delta T and add to the normal boiling point to find the new boiling point.
To find the boiling point of a solution, you can use the formula for boiling point elevation:
ΔTb = Kbp * msolute
Where:
ΔTb is the boiling point elevation
Kbp is the molal boiling point elevation constant
msolute is the molality of the solute (moles of solute per kg of solvent)
To solve this problem, we'll need to follow these steps:
Step 1: Calculate the molality (msolute) of the solute.
Given: 0.2 moles of solute and 125 grams of benzene.
To calculate the molality, first convert grams of benzene to kilograms:
mass of benzene = 125 grams = 0.125 kg
Then, calculate the molality using the formula:
msolute = moles of solute / mass of solvent (in kg)
msolute = 0.2 moles / 0.125 kg
Step 2: Calculate the boiling point elevation (ΔTb).
Given: Kbp = +2.53 °C/m and msolute from Step 1.
Using the formula:
ΔTb = Kbp * msolute
Step 3: Calculate the boiling point of the solution.
Given: Pure benzene has a boiling point of 80.1 °C.
The boiling point of the solution is given by:
Boiling point of solution = Boiling point of pure solvent + ΔTb
Substituting the values:
Boiling point of solution = 80.1 °C + ΔTb
Now, calculate the boiling point of the solution by substituting the values of ΔTb from Step 2 into the equation.
I hope this explanation helps in solving the problem!