Question

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

a. adding $10gofLiCl$to role="math" localid="1652764006746" $15gof{H}_{2}O$

b. adding $25gofLiCl$to $50gof{H}_{2}O$

c. adding $75gofLiCl$to $150gof{H}_{2}O$

Short answer

(Part a) The mixture of adding $10gofLiCl$to $15gof{H}_{2}O$forms Super-saturated solution.

(Part b) The mixture of adding $25gofLiCl$to $50gof{H}_{2}O$forms Un-saturated solution.

(Part c) The mixture of adding $75gofLiCl$to $150gof{H}_{2}O$forms Un-saturated solution.

Step-by-step solution

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.

Given Information (Part a).

Given:

Adding$10gofLiCl$to$15gof{H}_{2}O$

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

localid="1653898265752" $$\begin{gathered} =\frac{55\mathrm{g}}{100mL} \\ =0.55gm{L}^{-1} \end{gathered}$$

Explanation (part a).

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

localid="1653898286267" $$\begin{gathered} =\frac{15\mathrm{g}}{10\mathrm{mL}} \\ =1.5\mathrm{g}{\mathrm{mL}}^{-1} \end{gathered}$$

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

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

Given Information (step b).

Given:

Adding$25gofLiCl$to$50gof{H}_{2}O$

The solubility of $LiClat{25}^{\circ }C$

localid="1653898302295" $$\begin{gathered} =\frac{55\mathrm{g}}{100\mathrm{mL}} \\ =0.55g{\mathrm{mL}}^{-1} \end{gathered}$$

Explanation (part b).

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

localid="1653898322858" $$\begin{gathered} =\frac{25\mathrm{g}}{50\mathrm{mL}} \\ =0.5\mathrm{g}{\mathrm{mL}}^{-1} \end{gathered}$$

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

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

Given Information (step c).

Given:

Adding$75gofLiCl$to$150gof{H}_{2}O$

The solubility of $LiClat{25}^{\circ }C$

localid="1653898338065" $$\begin{gathered} =\frac{55\mathrm{g}}{100\mathrm{mL}} \\ =0.55g{\mathrm{mL}}^{-1} \end{gathered}$$

Explanation (part c).

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

localid="1653898358725" $$\begin{gathered} =\frac{75\mathrm{g}}{150\mathrm{mL}} \\ =0.5\mathrm{g}{\mathrm{mL}}^{-1} \end{gathered}$$

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

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