to what volume should you dilute 125mL of an 8.00M CuCl2 solution so that 50.0mL of the diluted solution contains 5.9g KNO3
I have a hard time determing the
M1V1=M2V2 is
Would you like to check your post please? You can't dilute CuCl2 solution and expect KNO3 at the end.
To determine the volume of the diluted solution, we can use the formula M1V1 = M2V2, where M1 is the initial molar concentration, V1 is the initial volume, M2 is the final molar concentration, and V2 is the final volume.
First, let's identify the known values:
M1 = 8.00 M (initial molar concentration of CuCl2 solution)
V1 = 125 mL (initial volume of the CuCl2 solution)
M2 = unknown (final molar concentration of the diluted solution)
V2 = 50.0 mL (final volume of the diluted solution)
We need to solve for M2, so rearrange the equation:
M2 = (M1 * V1) / V2
Now, substitute the known values into the equation:
M2 = (8.00 M * 125 mL) / 50.0 mL
First, convert the initial volume (V1) and the final volume (V2) to liters:
V1 = 0.125 L
V2 = 0.050 L
Substitute these values into the equation:
M2 = (8.00 M * 0.125 L) / 0.050 L
M2 = (1.000 mol/L) / 0.050 L
M2 = 20.00 M
The final molar concentration of the diluted solution is 20.00 M.
Now, regarding your question about the formula M1V1 = M2V2, it is known as the dilution formula. This formula is derived from the principle of the conservation of moles (or mass). In a dilution, the number of moles (or mass) before dilution is the same as the number of moles (or mass) after dilution. This formula allows you to calculate either the initial or final concentration or volume of a solution when the other values are known.