How would you prepare 300 mL of .75M NaBr solution using 2.0M NaBr stock?

If i could get help on this it would be really appriciated!

Answer:
M1 x V1 = M2 x V2

(300 mL) x (0.75 M) = (V2) x (2.0 M)
225 mL*M = 2.0 M (V2)
225 mL*M / 2.0 M = V2
112.5 mL = V2

You would use 112.5 mL of 2.0 M NaBr stock solution and dilute it with distilled water to a volume of 300 mL
Use the dilution formula:
M1 x V1 = M2 x V2

(300 mL) x (0.75 M) = (V2) x (2.0 M)
V2 =112.5 mL

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