Question

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

Answer 1

Approx. `50.0*mL`....

Explanation:

`"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*mL` volume of `0.500*mol*L^-1` `NaBr`...

`n_"sodium bromide"=200xx10^-3*Lxx0.500*mol*L^-1=0.10*mol.`

And `2.00*mol*L^-1` `NaBr(aq)` is available...so....

`(0.10*mol)/(2.00*mol*L^-1)xx1000*mL*L^-1=50.0*mL`.

Answer 2

`"50.0 mL"` of the `"2.00 M NaBr"` solution is needed to make `"200.0 mL"` of a `"0.500 M NaBr"` solution.

Explanation:

Use the dilution equation:

`C_1V_1=C_2V_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="2.00 M"`

`C_2="0.500 M"`

`V_2=200.0"mL"`

Unknown

`V_1`

Solution

Rearrange the equation to isolate `V_1`. Plug in the known values and solve.

`V_1=(C_2V_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)

`"50.0 mL"` of the `"2.00 M NaBr"` solution is needed to make `"200.0 mL"` of a `"0.500 M NaBr"` solution.