calculate the necessary amounts of solutes and solvent to make a 4 liter bottle of antifreeze that will have a 35 degree change in freezing point of water. solutes are Al2(SO4)3, KBr, Mg(NO3)2
delta T = i*Kf*molality
Solve for molality
molality = moles/kg solvent
solve for moles
moles = grams/molar mass
Solve for grams.
Go through the above for each solute.
i for Al2(SO4)3 = 5
i for KBr = 2
i for Mg(NO3)2 = 3
To calculate the necessary amounts of solutes and solvent for making a 4 liter bottle of antifreeze with a 35 degree change in freezing point of water, we need to follow a step-by-step process.
Step 1: Determine the freezing point depression constant (Kf) for water.
The value of Kf for water is 1.86 °C/m.
This constant represents the degree change in freezing point for each mole of solute added per kilogram of solvent.
Step 2: Determine the molality (m) of the solution.
The molality (m) is the number of moles of solute (n) divided by the mass of the solvent (m) in kilograms.
Step 3: Calculate the moles of solute required to achieve the desired freezing point depression.
The moles of solute (n) can be calculated using the formula:
n = m * molar mass of solute
where the molar mass can be found in the periodic table.
Step 4: Convert moles of solute to grams to find the amount required for each solute.
To convert moles to grams, multiply the number of moles by the molar mass of each solute.
Step 5: Determine the amount of solvent required.
The volume of the solvent (V) is given as 4 liters.
Now, let's apply these steps to the given solutes:
1. Al2(SO4)3:
- Find the molar mass of Al2(SO4)3:
2(27.0 g/mol Al) + 3(32.1 g/mol S) + 12(16.0 g/mol O) = 342.2 g/mol
- Calculate moles of Al2(SO4)3 required using the desired molality:
n = m * molar mass = (35 °C / 1.86 °C/mol) / 1000 g = 18.8 mol
- Convert moles to grams:
18.8 mol * 342.2 g/mol = 6473 g of Al2(SO4)3
2. KBr:
- Find the molar mass of KBr:
39.1 g/mol K + 79.9 g/mol Br = 119.0 g/mol
- Calculate moles of KBr required using the desired molality:
n = m * molar mass = (35 °C / 1.86 °C/mol) / 1000 g = 18.8 mol
- Convert moles to grams:
18.8 mol * 119.0 g/mol = 2237 g of KBr
3. Mg(NO3)2:
- Find the molar mass of Mg(NO3)2:
24.3 g/mol Mg + 2(14.0 g/mol N) + 6(16.0 g/mol O) = 148.3 g/mol
- Calculate moles of Mg(NO3)2 required using the desired molality:
n = m * molar mass = (35 °C / 1.86 °C/mol) / 1000 g = 18.8 mol
- Convert moles to grams:
18.8 mol * 148.3 g/mol = 2783 g of Mg(NO3)2
Therefore, you would need approximately 6473 grams of Al2(SO4)3, 2237 grams of KBr, and 2783 grams of Mg(NO3)2 to make a 4 liter bottle of antifreeze with a 35 degree change in freezing point of water.