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.

Short 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$72\mathrm{mL}$.

Step-by-step solution

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 $V_2$:

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