Calculate the molarity (M) of AlCl3 in a solution prepared by adding 0.750 L of 0.7 M AlCl3 and 0.250 L of 0.1 M AlCl3 . Assume volumes are additive

total moles: .7*.750+.250*.1

total volume= 1liter

Molalrity= total moles/1

0.9

To calculate the molarity (M) of AlCl3 in the solution, you need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's calculate the moles of AlCl3 from each solution.

For the first solution:
Molarity (M1) = 0.7 M (given)
Volume (V1) = 0.750 L (given)

Using the formula for moles:
moles (n1) = M1 * V1
moles (n1) = 0.7 M * 0.750 L = 0.525 moles of AlCl3

For the second solution:
Molarity (M2) = 0.1 M (given)
Volume (V2) = 0.250 L (given)

Using the formula for moles:
moles (n2) = M2 * V2
moles (n2) = 0.1 M * 0.250 L = 0.025 moles of AlCl3

Since the volumes are additive, the total volume of the solution is:
Total volume (Vt) = V1 + V2
Total volume (Vt) = 0.750 L + 0.250 L = 1.000 L

Now, add the moles from each solution to get the total moles:
Total moles (nt) = n1 + n2
Total moles (nt) = 0.525 moles + 0.025 moles = 0.550 moles of AlCl3

Finally, use the formula for molarity to calculate the molarity (M) of AlCl3 in the solution:
Molarity (M) = nt / Vt
Molarity (M) = 0.550 moles / 1.000 L = 0.550 M

Therefore, the molarity (M) of AlCl3 in the solution is 0.550 M.