How much ethyl alcohol, C2H5OH must be added to 500mL of water so that the solution will freeze at -10C? KfH2O = 1.86 C/molal
To determine the amount of ethyl alcohol (C2H5OH) needed to be added to 500 mL of water in order to freeze at -10°C, we can use the concept of freezing point depression and the formula:
ΔTf = Kf * m
where:
ΔTf is the change in freezing point
Kf is the molal freezing point depression constant for the solvent
m is the molality of the solution
First, let's find the change in freezing point (ΔTf). Since the water will freeze at 0°C, and we want it to freeze at -10°C, the change in freezing point can be calculated as:
ΔTf = -10°C - 0°C = -10°C
Next, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent, so we need to convert the volume of water (V) into mass (m).
Given that the density of water is approximately 1 g/mL, the mass (m) of 500 mL of water is:
m = V * density = 500 mL * 1 g/mL = 500 g
Now, let's calculate the molality (m) using the following equation:
m = n solute / mass solvent
Since we are adding ethyl alcohol C2H5OH as the solute to water, we can calculate the number of moles of C2H5OH (n solute) using its molar mass and the equation:
n solute = mass solute / molar mass C2H5OH
The molar mass of C2H5OH is 46.07 g/mol. Let's assume we want to find the amount of C2H5OH needed, expressed as moles (n).
Now, we can substitute the mass and molar mass values into the equation to find the number of moles of C2H5OH:
n = mass / molar mass
n = 500 g / 46.07 g/mol
Let's calculate n:
n ≈ 10.85 mol
Now, we have all the variables needed to calculate the molal freezing point depression constant (m), rearranging the formula:
ΔTf = Kf * m
we can solve for m:
m = ΔTf / Kf
m = -10°C / 1.86°C/molal
m ≈ -5.38 molal
Since the molality (m) is negative, indicating that ethyl alcohol decreases the freezing point of water, we need to add enough ethyl alcohol to achieve a molality of -5.38 molal.
Please note that the negative molality indicates a hypothetical scenario, as adding such a large amount of ethyl alcohol is not practical or common. The purpose of the calculation is to demonstrate the concept and mathematical process of freezing point depression.
delta T = K*m
substitute and solve for m
m = mols/kg solvent. You have kg solvent (500 mL = 500 g if density is 1.0 g/mL, and 500 g = 0.500 kg.) and m, solve for mols solute.
Then mols solute = grams/molar mass. You have molar mass and mols, solve for grams C2H5OH.