What volume of 9.54M sodium chloride solution must be added to the correct amount of water to produce 750. mL of a .327 M sodium chloride solution?
is it? M1V1=M2V2?
The solution you want contains
750 x 0.327 mmoles of NaCl
if the volume you add is x ml then the number of mmoles added is
9.54x mmoles = 750 x 0.327 mmoles
hence find x
the final volume is x+y=750 ml, where y is the volume of water added.
Two points....
9.54M NaCl would contain
9.54 x (23+35.5) g NaCl L^-1
9.54 x 58.4 g L^-1 = 557 g L^-1
as the max solubility of NaCl in water is 391 g L^-1, the question is not possible.
Also please remember to include the leading zero with decimals, so 0.327 M rather than .327 M, because the . can be easily missed.
To find the volume of the 9.54M sodium chloride solution needed, we can use the dilution formula:
C1V1 = C2V2
where:
C1 = initial concentration of the solution (9.54M)
V1 = initial volume of solution to be added (unknown)
C2 = final concentration of the solution (0.327M)
V2 = final volume of the solution (750 mL)
Let's rearrange the formula to find V1:
V1 = (C2 * V2) / C1
Now we can plug in the values and calculate:
V1 = (0.327M * 750 mL) / 9.54M
V1 = 25.70 mL
Therefore, you would need to add 25.70 mL of the 9.54M sodium chloride solution to the correct amount of water to produce 750 mL of a 0.327M sodium chloride solution.