Q. 9.114 - An Introduction to General, Organic, and Biological Chemistry

Question

In a laboratory experiment, a 15.0-mL sample of $\mathrm{KCl}$ solution is poured into an evaporating dish with a mass of 24.10 g.

The combined mass of the evaporating dish and $\mathrm{KCl}$ solution is 41.50 g . After heating, the evaporating dish and dry $\mathrm{KCl}$ have a combined mass of 28.28 g.

a. What is the mass percent $(\mathrm{m}/\mathrm{m})$ of the $\mathrm{KCl}$ solution?

b. What is the molarity $\left(\mathrm{M}\right)$of the $\mathrm{KCl}$ solution?

c. If water is added to 10.0 mL of the initial $\mathrm{KCl}$ solution to give a final volume     of 60.0 mL, what is the molarity of the diluted $\mathrm{KCl}$ solution?

Step-by-Step Solution

Answer

a)     $24\% \mathrm{m}/\mathrm{m}$  is the necessary mass percent of the   $\mathrm{KCl}$ solution.

b)     The molarity of the needed is The   $\mathrm{KCl}$ text solution is 3.73 M .

c)     The diluted solution's needed molarity $\mathrm{KCl}$ , The text solution is 0.62M.

Step - 1 Introduction

•   The volume of the  $\mathrm{KCl}$ solution sample utilized in the experiment was =15.0 mL.
• The mass of an evaporating dish is =24.10 g.
• The mass of the evaporating dish and the $\mathrm{KCl}$ solution combined is 41.50 g.
• As a result, the mass of the $\mathrm{KCl}$ solution =(41.50-24.10) g=17.40 g
• dry  $\mathrm{KCl}$=28.28 g  is the combined mass of the dry evaporating dish and $\mathrm{KCl}$=28.28 g.
• As a result, dry mass $\mathrm{KCl}$=(28.28-24.10) g=4.18 g

Step - 2 Given information (part-a)

a)

   Using the following expression, calculate the mass percent.

  $\text{ Mass percent }(\mathrm{m}/\mathrm{m})=\frac{\text{ Mass of solute }}{\text{ Mass of solution }}\times 100\%$

Step - 3 Explanation (part-a)

According to the problem,

•  The mass of the solute $\mathrm{KCl}=$4.18 g
•  The mass of the solution =17.40 g

Therefore,

           $\text{ Mass percent }(\mathrm{m}/\mathrm{m})=\frac{\text{ Mass of solute }}{\text{ Mass of solution }}\times 100\%$

                                                 $=\frac{4.18g}{17.40g}\times 100\%$

                                                 $=24\%$

  $24\% \mathrm{m}/\mathrm{m}$ is the necessary mass percent of the $\mathrm{KCl}$ solution.

Step - 4 Given information (part-b)

What is the mass percent of the  $\mathrm{KCl}$ solution (m / m) :

Step - 5 Explanation (part-b)

$\text{ Moles of }\mathrm{KCl}=\frac{\text{ mass of }\mathrm{KCl}}{\text{ molar mass of }\mathrm{KCl}}$

                        $=\frac{4.18 \mathrm{g}}{74.55 \mathrm{g}/\mathrm{mol}}$

                        $=0.056 \mathrm{mol}$

•  The volume of $\mathrm{KCl}$ text solution is 15.0. 0.015 mL or 0.015  mL.

• Use the following formula to calculate the molarity.

             $\text{ Molarity of }\mathrm{KCl}=\frac{\text{ moles of }\mathrm{KCl}}{\text{ Volume of }\mathrm{KCl}}$

                                        $=\frac{0.056 \mathrm{mol}}{0.015 \mathrm{L}}$

                                        $=3.73\mathrm{M}$

• The molarity of the needed is The $\mathrm{KCl}$ text solution is 3.73 M

Step - 6 Given information (part- c)

Here,

        ${\mathrm{C}}_1$=concentration of the concentrated solution=3.73 M

        $V_1$= volume of the concentrated solution=10 mL

        ${\mathrm{C}}_2$= concentration of the diluted solution

         ${\mathrm{V}}_2$=volume of the diluted solution=60 mL

 To solve for the unknown number C2, rearrange the dilution expression.

Step - 7 Explanation (part-c)

• In accordance with the dilution expression:

        ${\mathrm{C}}_1\times {\mathrm{V}}_1={\mathrm{C}}_2\times {\mathrm{V}}_2$

       $\text{ Here, }$

                     $C_2=\frac{V_1\times C_1}{V_2}$

                       ${\mathrm{V}}_1=\frac{3.73\mathrm{M}\times 10 \mathrm{mL}}{60 \mathrm{mL}}$

                         $=0.62\mathrm{M}$

• The diluted solution's needed molarity KCl , The text solution is 0.62M.