How many grams of ethanol C2H5OH(1),must be added to 500.0 g of water to make a solution that freezes at0.00 F? The molal freezing point depression constant water is 1.86 C.kg/mole
(F-32)*5/9 = (0-32)*5/9 = -32*5/9 = about -17.8 but you need to confirm.
delta T = Kf*molality
17.78/1.86 = molality = ?
?molality = mols/kg solvent.
Plug in 0.500 for kg solvent and ?molality and solve for mols.
Then mols = grams/molar mass.
You know mols and molar mass, solve for grams.
Post your work if you get stuck.
I need help on knowing how to do this problem
To determine how many grams of ethanol (C2H5OH) must be added to 500.0 g of water to make a solution that freezes at 0.00 °F, we need to use the concept of molal freezing point depression.
The formula to calculate the freezing point depression (ΔTf) is given by:
ΔTf = Kf * m
Where:
ΔTf = change in freezing point (in °C)
Kf = molal freezing point depression constant (in °C.kg/mol)
m = molality of the solute (in mol/kg)
We need to convert the given temperature from Fahrenheit to Celsius before proceeding with the calculation.
0.00 °F = -17.78 °C
Now, let's find the molality (m) of the solution. The molality is calculated using the following formula:
m = (moles of solute) / (mass of solvent in kg)
First, we need to convert the given mass of water to kg:
Mass of water = 500.0 g = 0.5 kg
Next, we need to calculate the moles of water using its molar mass:
Molar mass of water (H2O) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Moles of water = (mass of water) / (molar mass of water)
= 0.5 kg / (18.02 g/mol)
= 0.0277 mol
The molality (m) can now be calculated:
m = (moles of solute) / (mass of solvent in kg)
m = (moles of ethanol) / (0.5 kg)
Now, we can rearrange the freezing point depression equation to solve for moles of solute:
moles of solute = ΔTf / Kf
Since ΔTf is 17.78 °C (the change in freezing point) and Kf is 1.86 °C.kg/mol, we can substitute the values:
moles of solute = 17.78 °C / (1.86 °C.kg/mol)
= 9.56 mol
Finally, we can calculate the mass of ethanol (C2H5OH) needed using its molar mass:
Molar mass of ethanol (C2H5OH) = 2(12.01 g/mol) + 6(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol
Mass of ethanol = (moles of ethanol) * (molar mass of ethanol)
= 9.56 mol * (46.07 g/mol)
≈ 441.83 g
Therefore, approximately 441.83 grams of ethanol (C2H5OH) must be added to 500.0 grams of water to make a solution that freezes at 0.00 °F.