You have a stock solution of 14.4 M NH3. how many mililiters of this solution should you dilute to make 1050mL of .220 M NH3
answer 16.o mL
Question if you take a 10.5mL portion of the stock solution and dilute it to a total volume of .600L, what will be the concentration of the final solution?
First part is mL x M = mL x M.
Second part is
14.4 x (10.5 mL/600 mL) = ??
0.1
14.8
To find the concentration of the final solution, you need to use the dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the stock solution (14.4 M NH3)
V1 = initial volume of the stock solution taken (10.5 mL)
C2 = final concentration of the diluted solution (unknown)
V2 = final volume of the diluted solution (0.600 L)
Rearranging the formula, we get:
C2 = (C1 * V1) / V2
Substituting the given values:
C2 = (14.4 M NH3 * 10.5 mL) / 0.600 L
Now, let's simplify the units:
C2 = (14.4 mol/L * 0.0105 L) / 0.600 L
C2 = 0.252 mol / 0.600 L
Finally, divide the number of moles by the volume to get the concentration:
C2 = 0.42 M NH3
Therefore, the concentration of the final solution after diluting a 10.5 mL portion of the stock solution to a total volume of 0.600 L will be 0.42 M NH3.