# How many mL of a 2.00 M NaBr solution are needed to make 200.0 mL of 0.500M NaBr?

May 13, 2018

Approx. $50.0 \cdot m L$....

#### Explanation:

$\text{Concentration"="Number of moles"/"Volume of solution}$...and so we use this quotient appropriately to find the THIRD value if the other two values are specified.

We want a $200 \cdot m L$ volume of $0.500 \cdot m o l \cdot {L}^{-} 1$ $N a B r$...

${n}_{\text{sodium bromide}} = 200 \times {10}^{-} 3 \cdot L \times 0.500 \cdot m o l \cdot {L}^{-} 1 = 0.10 \cdot m o l .$

And $2.00 \cdot m o l \cdot {L}^{-} 1$ $N a B r \left(a q\right)$ is available...so....

$\frac{0.10 \cdot m o l}{2.00 \cdot m o l \cdot {L}^{-} 1} \times 1000 \cdot m L \cdot {L}^{-} 1 = 50.0 \cdot m L$.

May 13, 2018

$\text{50.0 mL}$ of the $\text{2.00 M NaBr}$ solution is needed to make $\text{200.0 mL}$ of a $\text{0.500 M NaBr}$ solution.

#### Explanation:

Use the dilution equation:

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$,

where:

${C}_{1}$ is the initial concentration, ${V}_{1}$ is the initial volume, ${C}_{2}$ is the final concentration, and ${V}_{2}$ is the final volume.

Known

${C}_{1} = \text{2.00 M}$

${C}_{2} = \text{0.500 M}$

${V}_{2} = 200.0 \text{mL}$

Unknown

${V}_{1}$

Solution

Rearrange the equation to isolate ${V}_{1}$. Plug in the known values and solve.

${V}_{1} = \frac{{C}_{2} {V}_{2}}{C} _ 1$

V_1=(0.500color(red)cancel(color(black)("M"))xx"200.0 mL" )/(0.200color(red)cancel(color(black)("M")))= "50.0 mL" (rounded to three significant figures)

$\text{50.0 mL}$ of the $\text{2.00 M NaBr}$ solution is needed to make $\text{200.0 mL}$ of a $\text{0.500 M NaBr}$ solution.