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

2 Answers
May 13, 2018

Answer:

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#.

May 13, 2018

Answer:

#"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.