Question

Calculate the final concentration of each of the following:

a. $1.0\mathrm{L}\mathrm{of}\mathrm{a}4.0\mathrm{M}{\mathrm{HNO}}_3$solutions is added to the water so that the final volume is $8.0\mathrm{L}$.

b. Water is added to $0.25\mathrm{L}\mathrm{of}\mathrm{a}6.0\mathrm{M}\mathrm{NaF}$the solution to make$2.0\mathrm{L}$of a diluted $\mathrm{NaF}$solution.

c. A role="math" localid="1653298318835" $50.0‐\mathrm{mL}$sample of a $8.0\%(\mathrm{m}/\mathrm{v})\mathrm{KBr}$solution is diluted with water so that the final volume is $200.0\mathrm{mL}$.

d. A $5.0‐\mathrm{mL}$sample of an$50.0\%(\mathrm{m}/\mathrm{v})$acetic acid $\left({\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\right)$solution is added to water to give a final volume of $25\mathrm{mL}$.

Short answer

a. The final concentration of $1.0\mathrm{L}\mathrm{of}\mathrm{a}4.0\mathrm{M}{\mathrm{HNO}}_3$solutions is added to the water so that the final volume is $8\mathrm{L}$is $0.5\mathrm{M}$.

b. The final concentration when water is added to $0.25\mathrm{L}\mathrm{of}\mathrm{a}6.0\mathrm{M}\mathrm{NaF}$ the solution to make $2.0\mathrm{L}$of a diluted $\mathrm{NaF}$solution is role="math" localid="1653300669777" $0.75\mathrm{M}$.

c. The final concentration of $50.0‐\mathrm{mL}$ the sample of a $8.0\%(\mathrm{m}/\mathrm{v})\mathrm{KBr}$solution is diluted with water so that the final volume is $200.0\mathrm{mL}$is $2\%$.

d. The final concentration of $5.0‐\mathrm{mL}$ sample of an $50.0\%(\mathrm{m}/\mathrm{v})$acetic acid $\left({\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\right)$solution is added to water to give a final volume of $25\mathrm{mL}$is $10\%$.

Step-by-step solution

Part (a) step 1: Given Information 

We need to calculate the final concentration of the $1.0\mathrm{L}\mathrm{of}\mathrm{a}4.0\mathrm{M}{\mathrm{HNO}}_3$solutions is added to the water so that the final volume is$8.0\mathrm{L}$.

Part (a) step 2: Explanation

We know

molarity $\left(\mathrm{M}\right)$$=4\mathrm{M}$

Litre is solution $=1\mathrm{L}$

The moles of solute = $1\times 4=4\text{moles}$

The final volume of solution $=8\mathrm{L}$

Molarity $\left(\mathrm{M}\right)=\frac{4\mathrm{mole}}{8\mathrm{L}}=0.5\mathrm{M}$

Part (b) step 1: Given Information 

We need to calculate the final concentration when water is added to$0.25\mathrm{L}\mathrm{of}\mathrm{a}6.0\mathrm{M}\mathrm{NaF}$the solution to make$2.0\mathrm{L}$of a diluted $\mathrm{NaF}$solution.

Part (b) step 2: Explanation

We know,

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

Litre is solution $=0.25\mathrm{L}$

The moles of solute =$6\times 0.25=1.5\text{moles}$

The final volume of solution $=2\mathrm{L}$

Molarity $\left(\mathrm{M}\right)=\frac{1.5\mathrm{mole}}{6\mathrm{L}}=0.75\mathrm{M}$

Part (c) step 1: Given Information 

We need to calculate the final concentration of $50.0‐\mathrm{mL}$ sample of a $8.0\%(\mathrm{m}/\mathrm{v})\mathrm{KBr}$solution is diluted with water so that the final volume is $200.0\mathrm{mL}$.

Part (c) step 2: Explanation

We know,

initial volume $=50\mathrm{mL}=0.05\mathrm{L}$

$$\begin{gathered} \%\mathrm{mass}(\mathrm{m}/\mathrm{v})=8\% \\ \mathrm{mass}\mathrm{of}\mathrm{solute}=\frac{\%\mathrm{mass}}{100}\times \left(\mathrm{solute}\mathrm{volume}\right) \\ =\frac{8}{100}\times 0.05=00.4\mathrm{g} \end{gathered}$$

Final volume of solution $=200mL=0.2L$

New $\%\mathrm{mass}(\mathrm{m}/\mathrm{v})=\frac{0.004\mathrm{g}}{0.2}\times 100=2\%$

Part (d) step 1: Given Information 

We need to calculate the final concentration of $5.0‐\mathrm{mL}$the sample of an$50.0\%(\mathrm{m}/\mathrm{v})$
an acetic acid $\left({\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\right)$solution is added to water to give a final volume of$25\mathrm{mL}$ .

Part (d) step 2: Explanation

We know,

initial volume $=5\mathrm{mL}=0.005\mathrm{L}$

$$\begin{gathered} \%\mathrm{mass}(\mathrm{m}/\mathrm{v})=50\% \\ \mathrm{mass}\mathrm{of}\mathrm{solute}=\frac{\%\mathrm{mass}}{100}\times \left(\mathrm{solute}\mathrm{volume}\right) \\ =\frac{50}{100}\times 0.005=0.0025\mathrm{g} \end{gathered}$$

Final volume of solution is $=25\mathrm{mL}=0.025\mathrm{L}$

New, $\%\mathrm{mass}(\mathrm{m}/\mathrm{v})=\frac{0.0025\mathrm{g}}{0.025}\times 100=10\%$