How many mLs of a stock solution of 1 M NH4NO3 do you need to get 0.001 M of Nitrogen in 22 liters of water?
1M stock in NH4NO3 is ?M in N? Must be 2M since there are two N in 1 mole NH4NO3.
Then mLsoln1 x Msoln1 = mL soln2 x Msoln2.
Plug in what you have and solve for the 1 unknown you have.
To calculate the volume of the stock solution needed, we can use the equation:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution needed
C2 = desired concentration of Nitrogen
V2 = volume of water
Given:
C1 = 1 M (concentration of the stock solution)
C2 = 0.001 M (desired concentration of Nitrogen)
V2 = 22 liters (volume of water)
To find V1 (volume of the stock solution needed), we rearrange the equation:
V1 = (C2 * V2) / C1
Now, let's plug in the values:
V1 = (0.001 M * 22 liters) / 1 M
V1 = 0.022 liters = 22 mL (since there are 1000 mL in a liter)
Therefore, you would need 22 mL of the 1 M stock solution of NH4NO3 to get a concentration of 0.001 M of Nitrogen in 22 liters of water.