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?

Short 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-by-step solution

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.