Question

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

a. a $1.5\mathrm{M}\mathrm{HCl}$solution prepared from$20.0\mathrm{mL}\mathrm{of}\mathrm{a}6.0\mathrm{M}\mathrm{HCl}$solution.

b. a $2.0\%(\mathrm{m}/\mathrm{v})\mathrm{LiCl}$solution prepared from $50.0\mathrm{mL}\mathrm{of}\mathrm{a}10.0\%(\mathrm{m}/\mathrm{v})\mathrm{LiCl}$solution

c. a $0.500\mathrm{M}{\mathrm{H}}_3{\mathrm{PO}}_4$solution prepared from $50.0\mathrm{mL}\mathrm{of}\mathrm{a}6.00\mathrm{M}{\mathrm{H}}_3{\mathrm{PO}}_4$solution

d. a $5.0\%(\mathrm{m}/\mathrm{v})$glucose solution prepared from $75\mathrm{mL}\mathrm{of}\mathrm{a}12\%(\mathrm{m}/\mathrm{v})$glucose solution

Short answer

a. The final volume of a $1.5\mathrm{M}\mathrm{HCl}$solution prepared$20.0\mathrm{mL}\mathrm{of}\mathrm{a}6.0\mathrm{M}\mathrm{HCl}$is $80\mathrm{mL}$.

b. The final volume of a $2.0\%(\mathrm{m}/\mathrm{v})\mathrm{LiCl}$solution prepared $50.0\mathrm{mL}\mathrm{of}\mathrm{a}10.0\%(\mathrm{m}/\mathrm{v})\mathrm{LiCl}$is $250\mathrm{mL}$.

c. The final volume of a $0.500\mathrm{M}{\mathrm{H}}_3{\mathrm{PO}}_4$solution prepared $50.0\mathrm{mL}\mathrm{of}\mathrm{a}6.00\mathrm{M}{\mathrm{H}}_3{\mathrm{PO}}_4$is $600\mathrm{mL}$.

d. The final volume of a $5.0\%(\mathrm{m}/\mathrm{v})$glucose solution prepared $75\mathrm{mL}\mathrm{of}\mathrm{a}12\%(\mathrm{m}/\mathrm{v})$is $180\mathrm{mL}$.

Step-by-step solution

Part (a) Step 1: Given Information

We need to find the final volume of the solution $1.5\mathrm{M}\mathrm{HCl}$prepared from $20.0\mathrm{mL}\mathrm{of}\mathrm{a}6.0\mathrm{M}\mathrm{HCl}$.

Part (a) Step 2: Explanation

Here,

molarity$\left(\mathrm{M}\right)=6\mathrm{M}$

Litres is solution $=20\mathrm{mL}=0.02\mathrm{L}$

$$\begin{gathered} \mathrm{mole}\mathrm{of}\mathrm{solute}=(\mathrm{molarity}\times (\mathrm{liters}\mathrm{of}\mathrm{solution}\left)\right) \\ =6\times 0.02=0.12\mathrm{mole} \end{gathered}$$

The final volume of solution is,

Final molarity$=15\mathrm{M}$

localid="1653449437524" $$\begin{gathered} \mathrm{Liters}\mathrm{of}\mathrm{solution}=\frac{\mathrm{solute}\mathrm{moles}}{\mathrm{molarity}} \\ =\frac{0.12}{1.5}\times 100=0.08\mathrm{L} \\ =80\mathrm{mL} \end{gathered}$$

Part (b) Step 1: Given Information

We need to find the final volume of the solution $2.0\%(\mathrm{m}/\mathrm{v})\mathrm{LiCl}$preparedfrom$50.0\mathrm{mL}\mathrm{of}\mathrm{a}10.0\%(\mathrm{m}/\mathrm{v})\mathrm{LiCl}$.

Part (b) Step 2: Explanation

We know,

$C_1{V}_1=C_2{V}_2$

$$\begin{gathered} {\mathrm{C}}_1=2\% \\ {\mathrm{C}}_2=10\% \\ {\mathrm{V}}_1=50\mathrm{mL} \end{gathered}$$

Now, finding the value of $V_2$:

$$\begin{gathered} {\mathrm{V}}_2=\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{C}}_2} \\ =\frac{(10\%)(50\mathrm{mL})}{2\%} \\ =250\mathrm{mL} \end{gathered}$$

Part (c) Step 1: Given Information

We need to find the final volume of the solution $0.500\mathrm{M}{\mathrm{H}}_3{\mathrm{PO}}_4$prepared from$50.0\mathrm{mL}\mathrm{of}\mathrm{a}6.00\mathrm{M}{\mathrm{H}}_3{\mathrm{PO}}_4$.

Part (c) Step 2: Explanation

We know,

$C_1{V}_1=C_2{V}_2$

$$\begin{gathered} {\mathrm{C}}_1=6\mathrm{M} \\ {\mathrm{C}}_2=0.5\mathrm{M} \\ {\mathrm{V}}_1=50\mathrm{mL} \end{gathered}$$

Now, finding the value of ${\mathrm{V}}_2$:

$$\begin{gathered} {\mathrm{V}}_2=\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{C}}_2} \\ =\frac{\left(6\right)(50\mathrm{mL})}{0.5} \\ =600\mathrm{mL} \end{gathered}$$

Part (d) Step 1: Given Information

We need to find the final volume of the solution$5.0\%(\mathrm{m}/\mathrm{v})$glucose solution prepared from localid="1653450814447" $75\mathrm{mL}\mathrm{of}\mathrm{a}12\%(\mathrm{m}/\mathrm{v})$.

Part (d) Step 2: Explanation

We know,

$C_1{V}_1=C_2{V}_2$

$$\begin{gathered} {\mathrm{C}}_1=12\% \\ {\mathrm{C}}_2=5\% \\ {\mathrm{V}}_1=75\mathrm{mL} \end{gathered}$$

Now, finding the value of $V_2$:

$$\begin{gathered} {\mathrm{V}}_2=\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{C}}_2} \\ =\frac{(12\%)(75\mathrm{mL})}{5\%} \\ =180\mathrm{mL} \end{gathered}$$