Nitric acid, HNO3, is available commercially at a concentration of 16 M.
What volume would you use to prepare 770 mL of a 0.19 M solution?
0.77*0.19M = X*16M
Solve for X in litres.
Mathmate's answer is quite good (and correct); I just want to point out that one need not convert from mL to L to solve the problem. One can use
mL x M = mL x M so
770*0.19=X*16
Solve for X and the answer is in mL.
Thank you Dr. Bob.
The answer will then be automatically in the unit given in the question.
To find the volume of the 16 M nitric acid solution needed to prepare a 0.19 M solution, we can use the formula:
C1V1 = C2V2
Where:
C1 = Initial concentration of the solution (16 M)
V1 = Volume of the initial solution we want to find
C2 = Final concentration of the solution (0.19 M)
V2 = Final volume of the solution (770 mL)
Rearranging the formula to solve for V1, we have:
V1 = (C2 * V2) / C1
Plugging in the given values, we get:
V1 = (0.19 M * 770 mL) / 16 M
Calculating this expression:
V1 = 9.1125 mL
Therefore, you would need to use approximately 9.11 mL of the 16 M nitric acid solution to prepare 770 mL of a 0.19 M solution.