A 1.00M sucrose solution freezes at -1.8 degree Celsius. What is the freezing point of 1.00M glucose solution?
-1.8C
To determine the freezing point of a 1.00M glucose solution, we can use the concept of freezing point depression. Freezing point depression is a colligative property, which means that it depends only on the number of solute particles in a solution, regardless of their identity.
In this case, both sucrose and glucose are non-electrolytes, which means they do not dissociate into ions when dissolved in water. Therefore, the van 't Hoff factor for both sucrose and glucose is 1, since there is only one particle per formula unit.
The freezing point depression can be calculated using the equation:
ΔT = Kf * m
Where:
ΔT is the freezing point depression,
Kf is the cryoscopic constant (a constant characteristic of the solvent),
m is the molality of the solution.
For water, the cryoscopic constant (Kf) is 1.86 °C/m.
Given that the sucrose solution freezes at -1.8 °C, we can calculate the molality (m) of the sucrose solution using the freezing point depression equation:
ΔT = Kf * m
-1.8 °C = (1.86 °C/m) * m
Solving for m, we get:
m = -1.8 °C / 1.86 °C/m
m ≈ -0.9686 m
Since molality cannot be negative, we take the absolute value:
m ≈ 0.9686 m
Now, we can use the obtained molality for the glucose solution to find the freezing point depression:
ΔT = Kf * m
ΔT = 1.86 °C/m * 0.9686 m
ΔT ≈ 1.803 °C
To find the freezing point of the glucose solution, subtract the freezing point depression from the freezing point of the pure solvent (water):
Freezing point of glucose solution = 0 °C - 1.803 °C
Therefore, the freezing point of a 1.00M glucose solution would be approximately -1.803 °C.