To what volume should you dilute 53 mL of a 12 M stock HNO3
solution to obtain a 0.138 M HNO3
solution?
C1V1=C2V2
C1=12 M
V1=53 mL
C2=0.138 M
V2=???
Solve for V2:
V2=(C1*V1)/C2
this isnt correct.
Did you add 53mL to the answer for V2???? Some thinking is required on your part.....V2 is the amount of solvent needed to be added, not the volume you need to add solvent to achieve.
Did you add 53mL to the answer for V2???? Some thinking is required on your part.....V2 is the amount of solvent needed to be added, not the total volume after addition of the solvent.
To calculate the volume to which you should dilute the 53 mL of a 12 M stock HNO3 solution, you can use the dilution equation:
C1V1 = C2V2
Where:
C1 = initial concentration of the stock solution (12 M)
V1 = initial volume of the stock solution (53 mL)
C2 = final concentration of the diluted solution (0.138 M)
V2 = final volume of the diluted solution (unknown)
Rearranging the equation to solve for V2, we have:
V2 = (C1V1) / C2
Substituting the given values into the equation:
V2 = (12 M * 53 mL) / 0.138 M
Calculating this, we get:
V2 = 4,616.92 mL
Therefore, you need to dilute the 53 mL of the 12 M stock HNO3 solution to a final volume of approximately 4,616.92 mL (or 4.62 L) in order to obtain a 0.138 M HNO3 solution.