Which of the following solutions of strong electrolytes contain the largest number of ions:

100.0 mL of 0.30 M AlCl3,
50.0 mL of 0.60 M MgCl2,
200.0 mL of 0.40 M NaCl?

If you determine the volumes of these three ionic compounds, NaCl has the greatest volume.

AlCl3--Has four ions
MgCl2--has 3
NaCl-has 2

Salt can hold more Na+ to Cl- collisions. which form ionic compounds since there is a greater volume.

To determine which solution contains the largest number of ions, we need to calculate the number of ions produced by each strong electrolyte solution.

For AlCl3, it dissociates into one Al3+ ion and three Cl- ions:
AlCl3 -> Al3+ + 3 Cl-

For MgCl2, it dissociates into one Mg2+ ion and two Cl- ions:
MgCl2 -> Mg2+ + 2 Cl-

For NaCl, it dissociates into one Na+ ion and one Cl- ion:
NaCl -> Na+ + Cl-

To find the number of ions in each solution, we multiply the concentration of the electrolyte (in moles per liter) by the number of ions it produces.

For 100.0 mL of 0.30 M AlCl3:
Number of moles of AlCl3 = 0.30 mol/L x (100.0 mL / 1000 mL) = 0.030 mol
Number of Al3+ ions = 1 x 0.030 mol = 0.030
Number of Cl- ions = 3 x 0.030 mol = 0.090

For 50.0 mL of 0.60 M MgCl2:
Number of moles of MgCl2 = 0.60 mol/L x (50.0 mL / 1000 mL) = 0.030 mol
Number of Mg2+ ions = 1 x 0.030 mol = 0.030
Number of Cl- ions = 2 x 0.030 mol = 0.060

For 200.0 mL of 0.40 M NaCl:
Number of moles of NaCl = 0.40 mol/L x (200.0 mL / 1000 mL) = 0.080 mol
Number of Na+ ions = 1 x 0.080 mol = 0.080
Number of Cl- ions = 1 x 0.080 mol = 0.080

Comparing the number of ions in each solution:
- AlCl3 solution has 0.030 Al3+ ions and 0.090 Cl- ions.
- MgCl2 solution has 0.030 Mg2+ ions and 0.060 Cl- ions.
- NaCl solution has 0.080 Na+ ions and 0.080 Cl- ions.

From the analysis, we can conclude that the AlCl3 solution contains the largest number of ions, with 0.090 Cl- ions.

To determine which of the given solutions contains the largest number of ions, we need to consider the ionization of each compound. Strong electrolytes completely ionize in water, so we can calculate the number of ions in each solution by multiplying the concentration of the compound by the number of ions it dissociates into.

For AlCl3:
AlCl3 dissociates into 1 Al3+ ion and 3 Cl- ions. So, in 1 mole of AlCl3, we have a total of 4 moles of ions (1 Al3+ and 3 Cl-).
Therefore, in 0.30 M AlCl3, we have 0.30 x 4 = 1.20 moles of ions.

For MgCl2:
MgCl2 dissociates into 1 Mg2+ ion and 2 Cl- ions. So, in 1 mole of MgCl2, we have a total of 3 moles of ions (1 Mg2+ and 2 Cl-).
Therefore, in 0.60 M MgCl2, we have 0.60 x 3 = 1.80 moles of ions.

For NaCl:
NaCl dissociates into 1 Na+ ion and 1 Cl- ion. So, in 1 mole of NaCl, we have a total of 2 moles of ions (1 Na+ and 1 Cl-).
Therefore, in 0.40 M NaCl, we have 0.40 x 2 = 0.80 moles of ions.

Comparing the number of moles of ions in each solution:
AlCl3 (1.20 moles of ions) > MgCl2 (1.80 moles of ions) > NaCl (0.80 moles of ions)

Therefore, the solution of 50.0 mL of 0.60 M MgCl2 contains the largest number of ions.