what volume of .0250 M MgCl2 should be diluted to 250.0 mL to obtain a solution with [Cl]=.0135?
mL1 x M1 = mL2 x M2 BUT since MgCl2 contains two atoms Cl2/molecule, make the M of MgCl2 just twice that listed; i.e., as 0.0250*2 = 0.050 M in chloride.
To determine the volume of 0.0250 M MgCl2 that should be diluted to 250.0 mL to obtain a solution with [Cl]=0.0135, we can use the equation:
M1V1 = M2V2
Where:
M1 = initial molarity of MgCl2
V1 = initial volume of MgCl2
M2 = final molarity of Cl-
V2 = final volume of the diluted solution
First, let's rearrange the equation to solve for V1:
V1 = (M2 * V2) / M1
We are given:
M1 = 0.0250 M
V2 = 250.0 mL (or 0.250 L)
M2 = 0.0135 M
Plugging in these values, we can calculate V1:
V1 = (0.0135 M * 0.250 L) / 0.0250 M
V1 = 0.135 L
Therefore, 0.135 liters (or 135 milliliters) of 0.0250 M MgCl2 should be diluted to 250.0 mL to obtain a solution with [Cl-] = 0.0135.