What is the boiling point of an aqueous solution that freezes at -6.7 degress C. Please help.
What is 100+6.7 (1.86/.52) ?
To determine the boiling point of an aqueous solution, we need to use the concept of boiling point elevation. Boiling point elevation occurs because adding a solute (such as salt or sugar) to water raises its boiling point.
In this case, we are given that the aqueous solution freezes at -6.7 degrees Celsius, which means the freezing point has been lowered. The freezing point depression is a reflection of the molal concentration of the solute in the solution.
To find the boiling point elevation, we can use the equation:
ΔTb = Kbm
Where:
ΔTb is the boiling point elevation,
Kb is the boiling point elevation constant (specific to the solvent),
b is the molal concentration of the solute (in mol/kg),
m is the molality of the solution (in mol/kg).
Since we are given the freezing point depression instead of the molal concentration, we will also need the freezing point depression constant (Kf) to find the molality of the solution.
Once we have the molality of the solution, we can use the boiling point elevation constant (Kb) specific to water.
By rearranging the equation, we can solve for the boiling point elevation (ΔTb):
ΔTb = Kf × m
Now we need the freezing point depression constant (Kf) for the solvent, which is water. The Kf value for water is approximately 1.86 °C/m.
Using the formula, we have:
ΔTb = Kf × m
ΔTb = 1.86 °C/m × m
ΔTb = 1.86 °C
Therefore, the boiling point of the aqueous solution will be raised by 1.86 degrees Celsius. To find the boiling point, we add this value to the normal boiling point of water, which is 100 degrees Celsius.
Boiling point of aqueous solution = Normal boiling point of water + ΔTb
Boiling point of aqueous solution = 100 °C + 1.86 °C
Boiling point of aqueous solution = 101.86 °C
So, the boiling point of this aqueous solution is approximately 101.86 degrees Celsius.