Q. 9.57 - An Introduction to General, Organic, and Biological Chemistry

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

Step-by-Step Solution

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 $180 \mathrm{mL}$.

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$=15 \mathrm{M}$

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

Part (b) Step 1: Given Information

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

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  $75 \mathrm{mL} \mathrm{of} \mathrm{a} 12 \% (\mathrm{m}/\mathrm{v})$.

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 $V_2$:

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