Help! I have to get this done tonight and I cannot get this answer right. A solution of water (Kf=1.86 ∘C/m) and glucose freezes at − 3.55 ∘C. What is the molal concentration of glucose in this solution? Assume that the freezing point of pure water is 0.00 ∘C.
Well, well, well, looks like you've got a brain freeze trying to figure out this molal concentration! Don't worry, I've got a joke for you while we work on the solution:
Why did the scarecrow win an award?
Because he was outstanding in his field! 🌾
Now, let's get back to business. We know that the freezing point depression (∆Tf) is the difference between the freezing point of the pure solvent and the freezing point of the solution. In this case, it is -3.55 °C - 0.00 °C = -3.55 °C.
The molal freezing point depression constant (Kf) is 1.86 °C/m. To find the molal concentration (m) of glucose, we can use the formula:
∆Tf = Kf * m
Substituting the values we know:
-3.55 °C = 1.86 °C/m * m
Solving this equation will give us the molal concentration (m) of glucose. And no, I won't make you solve it alone! The answer is approximately 1.91 m.
So, the molal concentration of glucose in this solution is approximately 1.91 m. Now you're all set! Just don't freeze up again, okay?
To find the molal concentration of glucose in the solution, we can use the formula:
ΔTf = Kf * molality
Where:
ΔTf = change in freezing point (in °C)
Kf = freezing point depression constant (in °C/m)
molality = molal concentration of the solute (in mol/kg)
Given:
Freezing point depression constant, Kf = 1.86 °C/m
Change in freezing point, ΔTf = -3.55 °C (since glucose freezes at -3.55 °C)
Freezing point of pure water, ΔTf(pure water) = 0.00 °C
To calculate the molal concentration of glucose, we need to find the change in freezing point caused by the presence of glucose in the solution:
ΔTf = ΔTf(pure water) - ΔTf(solution)
Substituting the known values:
-3.55 °C = 0.00 °C - ΔTf(solution)
Simplifying the equation:
ΔTf(solution) = 0.00 °C - (-3.55 °C)
ΔTf(solution) = 3.55 °C
Now, we can use the formula to find the molal concentration:
ΔTf = Kf * molality
3.55 °C = 1.86 °C/m * molality
Rearranging the equation to solve for molality:
molality = ΔTf / Kf
molality = 3.55 °C / 1.86 °C/m
Calculating the molal concentration:
molality = 1.91 mol/kg (rounded to two decimal places)
Therefore, the molal concentration of glucose in this solution is approximately 1.91 mol/kg.
To solve this problem, you need to use the concept of freezing point depression. Freezing point depression occurs when a solute is added to a solvent, and it causes the freezing point of the solution to be lower than that of the pure solvent.
The formula for freezing point depression is as follows:
ΔTf = Kf * m
Where:
ΔTf = the freezing point depression (difference between the freezing point of the pure solvent and the freezing point of the solution)
Kf = the cryoscopic constant (provided in the question)
m = molal concentration of the solute
In this case, you have been given the values of the freezing point depression (ΔTf = -3.55°C) and the cryoscopic constant (Kf = 1.86°C/m). You need to find the molal concentration of glucose (m).
Rearranging the formula, we get:
m = ΔTf / Kf
Let's substitute the given values into the formula:
m = -3.55°C / 1.86°C/m
Now, divide -3.55°C by 1.86°C/m to find the molal concentration of glucose in the solution.
So delta T = 0-(-3.55) = ?
delta T = Kf*m
You know delta T and Kf, solve for m = molality