50 L of your 0.05M sodium. chloride solution, what volume of 1.25M solution would be necessary?
To do what?
Solution Prep'n by Dilution of Lab Stock ...
(Molarity x Volume)concentrated soln = (Molarity x Volume)diluted soln
M(conc) = 1.25M
V(conc) = ? (= volume needed to dilute to 50 Liters.)
M(dild) = 0.05M
V(dild) = 50 L
[(1.25M)(X) = (0.05M)(50L)
(x) = [(0.05M)(50L)]/(1.25M) = 2.0 L.
Transfer 2.0L of the 1.25M concentrated solution into mixing vessel and dilute with solvent (water) up to, but not to exceed 50L total volume.
To find the volume of a 1.25M sodium chloride solution that is necessary to make a 0.05M solution, you can use the equation:
(M1)(V1) = (M2)(V2)
where:
M1 = molarity of the first solution (1.25M)
V1 = volume of the first solution (unknown)
M2 = molarity of the second solution (0.05M)
V2 = volume of the second solution (50 L)
Rearranging the equation to solve for V1, we have:
V1 = (M2 * V2) / M1
Now, let's substitute the given values into the equation and calculate V1:
V1 = (0.05M * 50 L) / 1.25M
V1 = 2 L
Therefore, you would need 2 liters of a 1.25M sodium chloride solution to make a 0.05M solution using 50 liters of the 0.05M solution.