Q. 9.27 - An Introduction to General, Organic, and Biological Chemistry

Question

A solution containing $80 g of KCl$ in $200 g of H_{2}O$ at ${50}^{\circ }C$ is cooled to ${20}^{\circ }C$

a. How many grams of $KCl$ remain in solution at ${20}^{\circ }C$ ?

b. How many grams of solid $KCl$ crystallized after cooling?

Step-by-Step Solution

Answer

(part a) $68$ grams of $KCl$ remaining in the solution at ${20}^{\circ }C$

(part b) $12$ grams of solid $KCl$ crystalized after cooling.

Step1: Introduction (part a).

Cooling is the process of removing heat, that also generally leads to a lower temperature and/or phase change. Cooling can relate to every method of reducing a person's body temperature. Heat energy can be transferred such as through thermal radiation, heat conduction, or convection.

Step2: Given Information (part a).

The given information are used to find the solution:

$200 g of H_{2}O$ contains $80 g of KCl$

Temperatures is ${50}^{\circ }C to {20}^{\circ }C$

Step3: Find the grams of K C l at 20 ∘ C (part a).

At ${20}^{\circ }C$, the absorption of $KCl$ is $34 g of KCl$ in $100 g$ of water. This same volume of $KCl$ needed for $200 g$ of water is calculated as follows:

Amount of $KCl$,

$$\begin{gathered} =\frac{34 \mathrm{gKCl}}{100 {\mathrm{gH}}_2\mathrm{O}}\times 200\mathrm{g} {\mathrm{H}}_2\mathrm{O} \\ =68 \mathrm{g} \mathrm{KCl} \end{gathered}$$

Therefore, $68 g of KCl$ present in the solution at ${20}^{\circ }C$

Step4: Given Information (part b).

The information following are used to find the solution:

$200 g of H_{2}O$ contains $80 g of KCl$

Temperature is ${50}^{\circ }C to {20}^{\circ }C$

Step5: Find the grams of solid K C l after crystallization (part b).

To quantify the volume of solid $KCl$ crystalized after cooling, subtract this same substrate volume of $KCl$ requisite for $200 g$ of water from of the mass of $KCl$ in the solution.

$\text{ Mass of }\mathrm{KCl}\text{ for crystallized }=\text{ Mass of }\mathrm{KCl}\text{ in the solution }-\text{ Mass of }\mathrm{KCl}\text{ required at }20{}^{\circ }\mathrm{C}$

$$\begin{gathered} =80 \mathrm{g} \mathrm{KCl}-68 \mathrm{g} \mathrm{KCl} \\ =12 \mathrm{g} \mathrm{KCl} \end{gathered}$$