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

Question

Natalie Anton · Accepted 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.

Apply the equation for dilution: #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. Recognized #C_1="2.00 M"# #C_2="0.500 M"# #V_2=200.0"mL"# Not sure #V_1# Resolution 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.

Avery Adams · Answer

Approx. #50.0*mL#....

#"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#... The formula for "sodium bromide" is 200x10^-3*Lx0.500*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#.

Owen Ambrose · Answer

undefined To dilute a solution, you can use the dilution formula:

C1V1 = C2V2

Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume

Rearranging the formula to solve for V1:

V1 = (C2 * V2) / C1

Substituting the given values:

V1 = (0.500 M * 200.0 mL) / 2.00 M

V1 = 100.0 mL

So, 100.0 mL of the 2.00 M NaBr solution are needed to make 200.0 mL of 0.500 M NaBr.