We’re being asked to find the** ****freezing point** of a** 50.0- mL 1.54 % MgCl**_{2}** by mass solution** **after adding an additional 1.34 gMgCl**_{2} ( *i* = 2.5 for MgCl_{2}.)

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**:

$\overline{){\mathbf{\u2206}}{{\mathbf{T}}}_{{\mathbf{f}}}{\mathbf{=}}{\mathbf{i}}{\mathbf{\times}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{\times}}{\mathbf{m}}}$

∆T_{f} = change in freezing point = T_{f pure solvent} –T_{f solution}

K_{f} = 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 MgCl _{2} to solve for its freezing point by doing these steps: **

** **

*Step 1**. Determine the composition of the solution*

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

*$\overline{){\mathbf{density}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mass}}{\mathbf{volume}}}$*

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

$\overline{){\mathbf{m}}{\mathbf{a}}{\mathbf{s}}{\mathbf{s}}{\mathbf{}}{\mathbf{p}}{\mathbf{e}}{\mathbf{r}}{\mathbf{c}}{\mathbf{e}}{\mathbf{n}}{\mathbf{t}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{n}\mathbf{e}\mathbf{n}\mathbf{t}}{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{\times}}{\mathbf{100}}}$

**Step 2**. Calculate the **moles of the solute**.

**Step 3**. Calculate the **molality of the solution**.

$\overline{){\mathbf{molality}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mol}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}}$

Solute = MgCl_{2}

Solvent = Water (H_{2}O)

**Step 4**. Calculate the **freezing point of the s****olution**.

A 50.0-mL solution is initially 1.54% MgCl_{2} 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 MgCl_{2}? (Use *i* = 2.5 for MgCl_{2}.)

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Freezing Point Depression concept. If you need more Freezing Point Depression practice, you can also practice Freezing Point Depression practice problems.

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