What volume (in milliliters) of 0.355 M aqueous ammonia, NH3, could be made from 19.9 mL of concentrated aqueous ammonia (15.0 M)?
(atomic masses: N = 14.01, H = 1.008)
mL1 x M1 = mL2 x M2
Substitute.
mL1 x 0.355M = 19.9 mL x 15.0M
Solve for mL1
840.845
To find the volume of a 0.355 M solution of NH3 that can be made from 19.9 mL of a 15.0 M solution, we can use the concept of dilution.
The formula for dilution is:
M1V1 = M2V2,
where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
In this case:
M1 = 15.0 M (concentrated aqueous ammonia)
V1 = 19.9 mL (the initial volume of concentrated ammonia)
M2 = 0.355 M (desired final concentration of NH3)
V2 = ? (the final volume we are trying to find)
Plugging these values into the formula, we get:
(15.0 M)(19.9 mL) = (0.355 M)(V2)
Dividing both sides by 0.355 M, we find:
V2 = (15.0 M)(19.9 mL) / 0.355 M
Now we can calculate V2:
V2 ≈ 840 mL
Therefore, approximately 840 mL of 0.355 M aqueous ammonia can be made from 19.9 mL of concentrated aqueous ammonia.