# Problem: A 50.0-mL solution is initially 1.54% MgCl2 by mass and has a density of 1.05 g/mL.What is the freezing point of the solution after you add an additional 1.34 g MgCl2? (Use i = 2.5 for MgCl2.)

###### FREE Expert Solution

We’re being asked to find the freezing point of a 50.0- mL 1.54 % MgCl2 by mass solution after adding an additional 1.34 gMgCl2 ( i = 2.5 for MgCl2.)

Recall that the freezing point of a solution is lower than the freezing point of a pure solvent.

When calculating the freezing point of a solution, we’re going to use the equation for Freezing Point Depression:

$\overline{){\mathbf{∆}}{{\mathbf{T}}}_{{\mathbf{f}}}{\mathbf{=}}{\mathbf{i}}{\mathbf{×}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{×}}{\mathbf{m}}}$

∆Tf = change in freezing point = Tf pure solvent –Tf solution
Kf = freezing point depression constant
i = van' t Hoff factor of the solute = no. of ions
m = molality

We need to find the molality of the solution after adding additional MgCl2 to solve for its freezing point by doing these steps:

Step 1. Determine the composition of the solution after adding MgCl2.

Since we're given the volume and density of the solution, we can determine its mass using the equation:

Also, recall that Mass or weight percent (% by mass) is the percentage of a given element or compound within a solution.

The equation used to calculate for mass percent is shown below:

Mass component mass of solute
• Total mass  mass of solution

Step 2. Calculate the moles of the solute

Step 3. Calculate the molality of the solution.

Solute = MgCl­2
Solvent = Water (H2O)

Step 4. Calculate the freezing point of the solution. ###### Problem Details

A 50.0-mL solution is initially 1.54% MgCl2 by mass and has a density of 1.05 g/mL.

What is the freezing point of the solution after you add an additional 1.34 g MgCl2? (Use i = 2.5 for MgCl2.)