Q. 9.26 - An Introduction to General, Organic, and Biological Chemistry

Question

Determine whether each of the following solutions will be saturated or unsaturated at ${50}^{\circ }C$ :

a. adding $25 g$ of $KCl$ to $50 g of H_{2}O$

b. adding $150 g of NaN{O}_3$ to $75 g of H_{2}O$

c. adding $80 g$ of sugar to $25 g of H_{2}O$

Step-by-Step Solution

Answer

(part a) At ${50}^{\circ }C$, adding $25 g of KCl$ to $50 g of H_{2}O$. The solution is saturated.

(part b) At ${50}^{\circ }C$, adding $150 g of NaN{O}_3$ to $75 g of H_{2}O$. The solution is saturated.

(part c) At ${50}^{\circ }C$, adding $80 g$ of sugar to $25 g of H_{2}O$. The solution is saturated.

Step1: Introduction (part a).

A saturated sample with the full limit of saturated solution in it.

An unsaturated solution is one where the quantity of solute present in the solution is much less than the quantity that can be disintegrated at a given temperature.

The rate of dissolution of a product changes with temperature.

Step2: Given Information (part a).

Given: 

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

Step3: Explanation (part b).

At ${50}^{\circ }C,$ the rate of dissolution $KCl$ is $43 g/100 g H_{2}O.$

Determine the maximum amount of $KCl$ which can be disintegrated in $50 g of H_{2}O$ using the following conversion factor:

$$\begin{gathered} \text{= 50 }\mathrm{g} {\mathrm{H}}_2\mathrm{O}\times \frac{43 g \mathrm{KCl}}{100g H_2\mathrm{O}} \\ =21.5 \mathrm{g} \mathrm{KCl} \\ \approx 22 \mathrm{g} \mathrm{KCl} \end{gathered}$$

Therefore, $22 g of KCl$ dissolved in $50 g of H_{2}O.$

When $25 g of KCl$ is decided to add to $50 g of H_{2}O,$ the quantity of which could be dissolved in surpasses  the amount which can be added in $50 g of H_{2}O.$

As a result, adding $25 g of KCl$ to $50 g of H_{2}O$ outcomes in a saturated solution.

Step4: Given Information (part b).

Given:

Adding $150 g of NaN{O}_3$ to $75 g of H_{2}O.$

Step5: Explanation (part b).

At ${50}^{\circ }C,$ the dissolution rate of $NaN{O}_3$ is $114 g/100 g H_{2}O.$

Determine the maximum quantity of $NaN{O}_3$ that could be diluted in $75 g H_{2}O$ and use the following conversion factor:

$$\begin{gathered} =75 \mathrm{g} H_2\mathrm{O}\times \frac{114 \mathrm{g}}{100 \mathrm{g} H_{ 2}\mathrm{O}} \\ =85.5 \mathrm{g} {\mathrm{NaNO}}_3 \end{gathered}$$

Hence, $85.5 g of NaN{O}_3$ could indeed dissipate in $75 g of H_{2}O.$

Once is introduced to $75 g of H_{2}O,$the portion of $NaN{O}_3$ added is higher than the amount of $NaN{O}_3$ that can be dissolved in $75 g of H_{2}O.$

As a result, combining $150 g of NaN{O}_3$ with $75 g of H_{2}O$ yields a saturated solution.

Step6: Given Information (part c).

Given:

Adding $80 g$ of sugar to $25 g H_{2}O.$

Step7: Explanation (part c).

At ${50}^{\circ }C,$ the water content of sugar $\left(C_{12}{H}_{22}{O}_{11}\right)$ is $260 g/100 g H_{2}O.$

Find the theoretical sugar content that could be diluted in $25 g of H_{2}O$ using the following conversion factor:

$$\begin{gathered} =25g H_2\mathrm{O}\times \frac{260g C_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}}{100{ \mathrm{g} \mathrm{H}}_2\mathrm{O}} \\ =65 \mathrm{g} {\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11} \end{gathered}$$

Therefore, $65 g$ of sugar could indeed dissipate in $25 g H_{2}O.$

While$80 g$ of sugar is added to $25 g H_{2}O$, the amount of sugar added is higher than the amount of sugar which can be  solubilized in $25 g H_{2}O.$

As a result, attempting to add $80 g$ sugar $25 g H_{2}O$outcomes in a saturated solution.