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
To calculate the necessary amounts of solutes and solvent to make a 4 liter bottle of antifreeze, we need to consider the colligative properties of the solutes, specifically their effect on the freezing point of the solvent. In this case, we want the antifreeze to have a 35 degree change in freezing point of water.
Step 1: Determine the molecular weight of each solute:
- Al2(SO4)3:
- Al: 2 x 26.98 g/mol = 53.96 g/mol
- S: 3 x 32.06 g/mol = 96.18 g/mol
- O: 12 x 16.00 g/mol = 192.00 g/mol
- Total: 53.96 + 96.18 + 192.00 = 342.14 g/mol
- KBr:
- K: 39.10 g/mol
- Br: 79.90 g/mol
- Total: 39.10 + 79.90 = 119.00 g/mol
- Mg(NO3)2:
- Mg: 24.31 g/mol
- N: 2 x 14.01 g/mol = 28.02 g/mol
- O: 6 x 16.00 g/mol = 96.00 g/mol
- Total: 24.31 + 28.02 + 96.00 = 148.33 g/mol
Step 2: Calculate the molality of each solute required to achieve the desired freezing point depression:
- Molality (m) = Change in freezing point constant (Kf) / Molal freezing point depression constant
- For water, the molal freezing point depression constant (Kf) is approximately 1.86 °C/m.
- Therefore, the molality (m) for a 35 °C change in freezing point would be: m = (35 °C) / (1.86 °C/m) = 18.82 m
Step 3: Calculate the moles of each solute needed:
- Moles of solute (n) = molality (m) x solvent mass (kg)
- Since we have a 4 liter (4 kg) bottle of antifreeze, the solvent mass is 4 kg.
- Moles of Al2(SO4)3 = 18.82 m x 4 kg x (1 mol / 342.14 g) = 0.219 mol
- Moles of KBr = 18.82 m x 4 kg x (1 mol / 119.00 g) = 0.632 mol
- Moles of Mg(NO3)2 = 18.82 m x 4 kg x (1 mol / 148.33 g) = 0.507 mol
Step 4: Calculate the mass of each solute needed:
- Mass of solute = Moles of solute (n) x Molecular weight of solute
- Mass of Al2(SO4)3 = 0.219 mol x 342.14 g/mol = 74.83 g
- Mass of KBr = 0.632 mol x 119.00 g/mol = 75.13 g
- Mass of Mg(NO3)2 = 0.507 mol x 148.33 g/mol = 75.20 g
Therefore, to make a 4 liter bottle of antifreeze with a 35 degree change in freezing point of water, you would need approximately:
- 74.83 grams of Al2(SO4)3
- 75.13 grams of KBr
- 75.20 grams of Mg(NO3)2