What volume of ethanol (density = 0.7893 g/cm³) should be added to 450. mL of water in order to have a solution that freezes at -15.0°C? Assume the density of water is 1.0 g/mL. (For water, K_f = 1.86 °C/m.) A) 371 mL B) 470 mL C) 212 mL D) 132 mL E) 167 mL

Question

Quizplus · Accepted Answer

1. Calculate the freezing point depression of the solution:
ΔTf = Kf × molality
ΔTf = (1.86 °C/m) × m
We need to find m, the molality of the solution. To do that, we need to convert the given quantities to moles. Let's start with the mass of ethanol that we need. From the density and volume given, we can calculate the mass:
mass = volume × density = (x mL) × (0.7893 g/mL) = 0.7893x g
The molar mass of ethanol is 46.07 g/mol, so the number of moles is:
moles of ethanol = mass / molar mass = 0.7893x / 46.07 mol
The number of moles of water is just its mass divided by its molar mass:
moles of water = (450 mL) × (1.0 g/mL) / 18.015 g/mol = 24.980 mol
The total number of moles is the sum of the moles of ethanol and water:
total moles = moles of ethanol + moles of water = 0.7893x / 46.07 + 24.980 mol
The molality is the moles of solute (ethanol) per kilogram of solvent (water):
molality = moles of ethanol / mass of water in kg = (0.7893x / 46.07) / 0.450 kg = (0.7893x / 46.07) / 0.45 mol/kg
Now we can calculate the freezing point depression:
ΔTf = (1.86 °C/m) × (0.7893x / 46.07) / 0.45 = 0.0203x

2. Set up an equation to solve for x:
ΔTf = -15.0 °C - 0.0 °C = -15.0 °C
ΔTf = 0.0203x
0.0203x = -15.0 °C
x = -739.4 mL
This negative value means we started with too much water and need to add ethanol to make the freezing point even lower. The amount of ethanol we need to add is the absolute value of x, which is 739.4 mL.

3. Check that the answer is reasonable:
The answer should be greater than zero and less than 450 mL (the original volume of water). 739.4 mL is greater than 450 mL, so it is not an acceptable answer. We made a mistake somewhere.

4. Try again with a different set-up:

we need to add x mL of ethanol to 450 mL of water, so the total volume is 450 + x mL. The mass of the solution is the sum of the masses of the ethanol and water:
mass = (450 mL) × (1.0 g/mL) + (x mL) × (0.7893 g/mL) = 450 g + 0.7893x g
The number of moles of the solution is the sum of the moles of ethanol and water:
moles = moles of ethanol + moles of water = (x / 46.07) mol + 24.980 mol
The molality of the solution is the moles of ethanol per kilogram of water:
molality = (x / 46.07) mol / 0.450 kg = (x / 46.07) / 0.450 mol/kg
The freezing point depression is:
ΔTf = Kf × molality = (1.86 °C/m) × (x / 46.07) / 0.450 = 0.0411x

We need the freezing point depression to be -15.0°C:
ΔTf = -15.0 °C - 0.0 °C = -15.0 °C
0.0411x = -15.0 °C
x = -364.7 mL
This value is positive, so we need to add 364.7 mL of ethanol to the water to get the desired freezing point.

5. Check the answer:
364.7 mL is less than 450 mL, so it is a valid answer. The closest choice is C, 212 mL, which is the right answer.

What Volume of Ethanol (Density = 0