an aqueous solution freezes at -1.5C.Calculate the boiling point of the solution if the Kf is 1.86 and Kb is 0.51
I suppose we assume the solute is a non-volatile one and does not ionize.
dTfreeze = Kf*m
dT = 1.5, you know Kf, solve for molality.
Then dTboiling = Kb*m
You know Kb and m, solve for dT and add it to 100 C.
Post your work if you get stuck.
To calculate the boiling point of the solution, we can use the formula:
ΔTb = Kb * molality
Where:
ΔTb is the change in boiling point,
Kb is the boiling point elevation constant,
and molality is the molal concentration of the solution.
However, in order to use this formula, we need to know the molality of the solution. Since the question does not provide this information, we cannot calculate the boiling point accurately.
If you have any other information or data related to the solution, please provide it, and I will be happy to assist you further.
To calculate the boiling point of the solution, we need to use the equation:
ΔTb = Kb * m
where ΔTb is the boiling point elevation, Kb is the molal boiling point constant, and m is the molality of the solution.
Given that Kb = 0.51, we need to find the molality of the solution.
Since it is an aqueous solution, we can assume it is a solution of a solute dissolved in water. The solute could be any substance, but let's assume it is a nonvolatile solute such as a sugar or salt.
To find the molality (m), we need to know the amount of solute (in moles) and the mass of the solvent (in kilograms).
Since we don't have the specific values for the solute and solvent, we cannot directly calculate the molality (m).
However, we can use the formula for freezing point depression to find the molality first, and then use that to calculate the boiling point elevation.
The formula for freezing point depression is:
ΔTf = Kf * m
where ΔTf is the freezing point depression, Kf is the molal freezing point constant, and m is the molality of the solution.
Given that ΔTf = -1.5 °C and Kf = 1.86, we can rearrange the equation to solve for the molality (m):
m = ΔTf / Kf
m = -1.5 °C / 1.86
m ≈ -0.806 molal
Now that we have the molality (m = -0.806 molal), we can calculate the boiling point elevation (ΔTb) using the equation:
ΔTb = Kb * m
ΔTb = 0.51 * (-0.806)
ΔTb ≈ -0.411 °C
Since the aqueous solution freezes at -1.5 °C, the boiling point of the solution will be:
Boiling point = -1.5 °C + ΔTb
Boiling point = -1.5 °C + (-0.411 °C)
Boiling point ≈ -1.911 °C
Therefore, the boiling point of the solution is approximately -1.911 °C.