Task:
How many milliliters of water have to be added to 300 mL of stock solution of 11.1 M HNO3 to have final concentration of 6.00 M?
Solution:
This dilution problem uses the equation
С1V1 = C2V2;
C1 = 11.1 M - the initial molarity (concentration)
V1 = 300 mL - the initial volume
C2 = 6 M - the desired molarity (concentration)
V2 = (300 + x mL) - the volume of the desired solution
(11.1 M) (300 mL) = (6.00 M)(300 mL + x )
3330 M mL= 1800 M mL + 6.00x M
(3330-1800) = 6.00x
1530 = 6.00x
1530 / 6.00 = x
x = 255
255 mL needs to be added to the original 300 mL solution in order to dilute it from 11.1 M to 6.00 M.
Answer: 255 mL of water have to be added.
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