Q. 9.58 - An Introduction to General, Organic, and Biological Chemistry

Question

Determine the final volume, in milliliters, of each of the following: 

a. a $1.00 \% (\mathrm{m}/\mathrm{v}) {\mathrm{H}}_2{\mathrm{SO}}_4$solution prepared from $10 \mathrm{mL} \mathrm{of} \mathrm{a} 20.0 \% {\mathrm{H}}_2{\mathrm{SO}}_4$solution.

b. a $0.10 \mathrm{M} \mathrm{HCl}$solution prepared from $25 \mathrm{mL} \mathrm{of} \mathrm{a} 6.0 \mathrm{M} \mathrm{HCl}$solution.

c. a $1.0 \mathrm{M} \mathrm{NaOH}$ solution prepared from $50.0 \mathrm{mL} \mathrm{of} \mathrm{a} 12 \mathrm{M} \mathrm{NaOH}$solution.

d. a $1.0 \% (\mathrm{m}/\mathrm{v}) {\mathrm{CaCl}}_2$solution prepared from $18 \mathrm{mL} \mathrm{of} \mathrm{a} 4.0 \% (\mathrm{m}/\mathrm{v}) {\mathrm{CaCl}}_2$ solution.

Step-by-Step Solution

Answer

a. The final volume of a $1.00 \% (\mathrm{m}/\mathrm{v}) {\mathrm{H}}_2{\mathrm{SO}}_4$ solution prepared $10 \mathrm{mL} \mathrm{of} \mathrm{a} 20.0 \% {\mathrm{H}}_2{\mathrm{SO}}_4$is  $200 \mathrm{mL}$.

b. The final volume of a $0.10 \mathrm{M} \mathrm{HCl}$solution prepared $25 \mathrm{mL} \mathrm{of} \mathrm{a} 6.0 \mathrm{M} \mathrm{HCl}$is $1500 \mathrm{mL}$.

c. The final volume of a $1.0 \mathrm{M} \mathrm{NaOH}$ solution prepared $50.0 \mathrm{mL} \mathrm{of} \mathrm{a} 12 \mathrm{M} \mathrm{NaOH}$is $600 \mathrm{mL}$.

d. The final volume of a $1.0 \% (\mathrm{m}/\mathrm{v}) {\mathrm{CaCl}}_2$ solution prepared $18 \mathrm{mL} \mathrm{of} \mathrm{a} 4.0 \% (\mathrm{m}/\mathrm{v}) {\mathrm{CaCl}}_2$is $72 \mathrm{mL}$.

Part (a) Step 1: Given Information

We need to find the final volume of the solution $1.00 \% (\mathrm{m}/\mathrm{v}) {\mathrm{H}}_2{\mathrm{SO}}_4$ prepared from $10 \mathrm{mL} \mathrm{of} \mathrm{a} 20.0 \% {\mathrm{H}}_2{\mathrm{SO}}_4$. 

Part (a) Step 2: Explanation

The addition of solvent increases the volume, the amount of solute does not change. The same in the concentrated solution and the diluted solution. Moles or grams   of solute$=$Moles or grams   of solute

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

$$\begin{gathered} {\mathrm{C}}_1=20\% \\ {\mathrm{C}}_2=1 \% \\ {\mathrm{V}}_1=10 \mathrm{mL} \end{gathered}$$

Now, finding the value of $V_2$:

$$\begin{gathered} {\mathrm{V}}_2=\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{C}}_2} \\ =\frac{(20 \%) (10 \mathrm{mL})}{1 \%} \\ =200 \mathrm{mL} \end{gathered}$$ 

Part (b) Step 1: Given Information

We need to find the final volume of the solution $0.10 \mathrm{M} \mathrm{HCl}$ prepared from $25 \mathrm{mL} \mathrm{of} \mathrm{a} 6.0 \mathrm{M} \mathrm{HCl}$. 

Part (b) Step 2: Explanation

The addition of solvent increases the volume, the amount of solute does not change. The same in the concentrated solution and the diluted solution. Moles or grams   of solute$=$Moles or grams   of solute

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

$$\begin{gathered} {\mathrm{C}}_1=6 \mathrm{M} \\ {\mathrm{C}}_2=0.1 \mathrm{M} \\ {\mathrm{V}}_1=25 \mathrm{mL} \end{gathered}$$

Now, finding the value of  $V_2$:

$$\begin{gathered} {\mathrm{V}}_2=\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{C}}_2} \\ =\frac{(6 \mathrm{M}) (25 \mathrm{mL})}{0.1 \mathrm{M}} \\ =1500 \mathrm{mL} \end{gathered}$$

Part (c) Step 1: Given Information

We need to find the final volume of the solution $1.0 \mathrm{M} \mathrm{NaOH}$ prepared from $50.0 \mathrm{mL} \mathrm{of} \mathrm{a} 12 \mathrm{M} \mathrm{NaOH}$.  

Part (c) Step 2: Explanation

The addition of solvent increases the volume, the amount of solute does not change. The same in the concentrated solution and the diluted solution. Moles or grams   of solute$=$Moles or grams   of solute

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

$$\begin{gathered} {\mathrm{C}}_1=12 \mathrm{M} \\ {\mathrm{C}}_2=1 \mathrm{M} \\ {\mathrm{V}}_1=50 \mathrm{mL} \end{gathered}$$

Now, finding the value of $V_2$:

$$\begin{gathered} {\mathrm{V}}_2=\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{C}}_2} \\ =\frac{(12 \mathrm{M}) (50 \mathrm{mL})}{1 \mathrm{M}} \\ =600 \mathrm{mL} \end{gathered}$$

Part (d) Step 1: Given Information

We need to find the final volume of the solution $1.0 \% (\mathrm{m}/\mathrm{v}) {\mathrm{CaCl}}_2$ prepared from $18 \mathrm{mL} \mathrm{of} \mathrm{a} 4.0 \% (\mathrm{m}/\mathrm{v}) {\mathrm{CaCl}}_2$. 

Part (d) Step 2: Explanation

The addition of solvent increases the volume, the amount of solute does not change. The same in the concentrated solution and the diluted solution. Moles or grams   of solute$=$Moles or grams   of solute

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

$$\begin{gathered} {\mathrm{C}}_1=4 \% \\ {\mathrm{C}}_2=1 \% \\ {\mathrm{V}}_1=18 \mathrm{mL} \end{gathered}$$

Now, finding the value of $V_2$:

$$\begin{gathered} {\mathrm{V}}_2=\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{C}}_2} \\ =\frac{(4 \%) (18 \mathrm{mL})}{1\%} \\ =72 \mathrm{mL} \end{gathered}$$