Final answer:
Approximately 12.281 grams of magnesium chloride must be added to 180 grams of water to lower its freezing point to -4.00°C, given the known freezing point depression constant and the van 't Hoff factor for MgCl2.
Step-by-step explanation:
The student is asking how much magnesium chloride (MgCl2) needs to be added to 180 grams of water to lower its freezing point to -4.00°C. The freezing point depression constant of water (Kf) is 1.86°C/m, which tells us how much the freezing point will decrease per molal concentration of the solute. Since the density of water at 0°C is 0.999843 g/mL, we can approximate 180 g of water to be 180 mL, because the density is nearly 1 g/mL.
Using the freezing point depression equation ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van 't Hoff factor (which is 3 for MgCl2 since it dissociates into three ions: Mg2+ and 2 Cl−), Kf is the freezing point depression constant, and m is the molality of the solution, we can solve for the amount needed. The desired ΔTf is -4.00°C, thus:
ΔTf is the target freezing point depression, i.e., 4.00°C (since freezing point is lowered from 0.00°C to -4.00°C).
To calculate the molality (m), we rearrange the equation to m = ΔTf / (i * Kf).
Substitute the known values into the equation:
-4.00°C = (3)(1.86°C/m)(m)
Solving for m gives m = ΔTf / (i * Kf) = -4.00°C / (3 * 1.86°C/m)
approximately 0.716 mol/kg.
Now, we multiply the molality by the mass of water to find the moles of solute needed:
0.716 mol/kg * 0.180 kg = 0.129 mol MgCl2.
To find the mass of magnesium chloride required, we multiply the moles by its molar mass (95.21 g/mol):
0.129 mol * 95.21 g/mol = approximately 12.281 grams.
Therefore, to lower the freezing point of 180 grams of water to -4.00°C, the student would need to add approximately 12.281 grams of magnesium chloride.