Calculate the boiling point of a 4.35 m benzene solution in phenol. (phenol bp=181.75 degree C and Kb=304 degree C/m)
sry we have not gotten that far yet. wish i could help
its all good
To calculate the boiling point of a solution, we can use the equation:
ΔTb = Kb × m
where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant, and m is the molality of the solute.
Given:
Kb (benzene) = 304 °C/m
m (benzene in phenol) = 4.35 m
We can substitute the values into the equation to find the boiling point elevation:
ΔTb = 304 °C/m × 4.35 m
ΔTb ≈ 1322.4 °C
To find the boiling point of the solution, we need to add the boiling point elevation to the boiling point of pure phenol:
Boiling point of pure phenol = 181.75 °C
Boiling point of the solution = 181.75 °C + 1322.4 °C
Boiling point of the solution ≈ 1504.15 °C
Therefore, the boiling point of the 4.35 m benzene solution in phenol is approximately 1504.15 °C.
To calculate the boiling point of a solution, you need to use the equation:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation
Kb is the ebullioscopic constant
m is the molality of the solution (moles of solute per kilograms of solvent)
In this case, we're given the following information:
Kb = 304 degrees C/m (ebullioscopic constant for phenol)
m = 4.35 m (molality of the benzene solution in phenol)
To find the boiling point of the solution, we need to calculate ΔTb first. Let's do that:
ΔTb = Kb * m
ΔTb = 304 degrees C/m * 4.35 m
ΔTb = 1322.4 degrees C
Now, to find the boiling point, we add the boiling point elevation (ΔTb) to the boiling point of the solvent:
Boiling point of solution = Boiling point of solvent + ΔTb
Boiling point of solution = 181.75 degrees C + 1322.4 degrees C
Boiling point of solution = 1504.15 degrees C
Therefore, the boiling point of the 4.35 m benzene solution in phenol is approximately 1504.15 degrees Celsius.