How many milliliters of a 110mL solution of 1.5 M AlCl3 must be used to make 70mL of a solution that has a concentration of 0.2M Cl–?
To calculate the amount of the 1.5 M AlCl3 solution needed to make a 0.2 M Cl- solution, we can use the formula:
C1V1 = C2V2
where C1 = concentration of the original solution, V1 = volume of the original solution, C2 = concentration of the final solution, and V2 = volume of the final solution.
In this case, we know:
C1 = 1.5 M (concentration of the original AlCl3 solution)
V1 = ? (volume of the original AlCl3 solution)
C2 = 0.2 M (concentration of the final Cl- solution)
V2 = 70 mL (volume of the final solution)
By rearranging the formula, we can solve for V1:
V1 = (C2V2) / C1
Plugging in the given values:
V1 = (0.2 M * 70 mL) / 1.5 M
Simplifying the equation:
V1 = (0.04 mol/L * 0.07 L) / 1.5 mol/L
Calculating:
V1 = 0.0028 L
Finally, converting volume to milliliters:
V1 = 0.0028 L * 1000 mL/L
The amount of the 1.5 M AlCl3 solution needed to make 70 mL of a 0.2 M Cl- solution is 2.8 mL.
How many moles Cl^- do you want? That is M x L = 0.2M x 0.070 L = 0.014 moles.
Since there are 3 Cl atoms in 1 molecule of AlCl3, you will need 1/3 of that in moles AlCl3. 0.014/3 = 0.00467 moles AlCl3.
moles AlCl3 = M AlCl3 x L AlCl3. You know moles and you know M (1.5M), solve for L and convert to mL. I get an answer of about 3 mL but that isn't exact.