(dilute 2 mL of ammonium to 100mL with H2O) + (10 mL ammonium added to 10 mL H2O) is what? in v/v % in ppm?

Question

Dilute 2 mL of ammonium to 100mL with water, then separately mix 10 mL of ammonium with 10 mL of water. Next add the two solutions together? what is the final v/v% , what is the ppm ?

Answer 1

Assume the volume is additive during mixing.
Hence the total volume of ammonium: 12mL
The total volume of water: 110 mL
The total volume of the solution is: 122mL
Hence the volume percentage of ammonium:
12/122 = 9.8%
And 9.8% = 98k ppm in volume

Answer 2

The first thing you need to remember is: M1 * V1 = M2 * V2 and since we need M2, this gives (M1 * V1) / V2 = M2 So you need to make sure of your unit, the unit of M is mols / liter, so 56.00 mL is 0.056 L then (0.056 L * 2.79 M) / 2.00 L = 0.078 M