Q. 9.116 - An Introduction to General, Organic, and Biological Chemistry

Question

Lithium chloride has a solubility of $55 g of LiCl$ in $100 g of H_{2}O$ at ${25}^{\circ }C$. Determine if each of the following mixtures forms an unsaturated or saturated solution at ${25}^{\circ }C$:

a. adding $10 g of LiCl$ to $15 g of H_{2}O$

b. adding $25 g of LiCl$ to $50 g of H_{2}O$

c. adding $75 g of LiCl$ to $150 g of H_{2}O$

Step-by-Step Solution

Answer

(Part a) The mixture of adding $10 g of LiCl$ to $15 g of H_{2}O$ forms Super-saturated solution.

(Part b) The mixture of adding $25 g of LiCl$ to $50 g of H_{2}O$ forms Un-saturated solution.

(Part c) The mixture of adding $75 g of LiCl$ to $150 g of H_{2}O$ forms Un-saturated solution.

Step1: Introduction.(Part a)

A saturated solution is one that contains the highest concentration of liquid solute at equilibrium.

A solution with a low solute concentration than a saturated solution. It'll be capable of dissolving a greater amount of solute.

A solution with a higher levels of solutes solute than a saturated solution.

Step2: Given Information (Part a).

Given:

Adding $10 g of LiCl$ to $15 g of H_{2}O$

The solubility of $LiCl at {25}^{\circ }C$ is

$$\begin{gathered} =\frac{55 \mathrm{g}}{100 mL} \\ =0.55 g m{L}^{-1} \end{gathered}$$

Step3: Explanation (part a).

Find the quantity of $LiCl$ per milliliter in the solution is:

$$\begin{gathered} =\frac{15 \mathrm{g}}{10 \mathrm{mL}} \\ =1.5 \mathrm{g} {\mathrm{mL}}^{-1} \end{gathered}$$

This is greater than the solubility of $LiCl=0.55 g m{l}^{-1}$

As a result, this mixture forms the Super-saturate solution.

Step4: Given Information (step b).

Given:

Adding $25 g of LiCl$ to $50 g of H_{2}O$

The solubility of $LiCl at {25}^{\circ }C$

$$\begin{gathered} =\frac{55 \mathrm{g}}{100 \mathrm{mL}} \\ =0.55 g {\mathrm{mL}}^{-1} \end{gathered}$$

Step5: Explanation (part b).

Find the quantity of $LiCl$per milliliter in the solution is:

$$\begin{gathered} =\frac{25 \mathrm{g}}{50 \mathrm{mL}} \\ =0.5\mathrm{g} {\mathrm{mL}}^{-1} \end{gathered}$$

This is lesser than the solubility of $LiCl=0.55 g m{l}^{-1}$

As a result, this mixture forms the Un-saturate solution.

Step6: Given Information (step c).

Given:

Adding $75 g of LiCl$ to $150 g of H_{2}O$

The solubility of $LiCl at {25}^{\circ }C$

$$\begin{gathered} =\frac{55 \mathrm{g}}{100 \mathrm{mL}} \\ =0.55 g {\mathrm{mL}}^{-1} \end{gathered}$$

Step7: Explanation (part c).

Find the quantity of $LiCl$ per milliliter in the solution is:

$$\begin{gathered} =\frac{75 \mathrm{g}}{150 \mathrm{mL}} \\ =0.5 \mathrm{g} {\mathrm{mL}}^{-1} \end{gathered}$$

This is lesser than the solubility of $LiCl=0.55 g m{l}^{-1}$

As a result, this mixture forms the Un-saturate solution.