How many number of ions of each type are present in the following aqueous solution?

113 mL of .490 M aluminum chloride.

I answered this below.

55370

To determine the number of ions of each type present in the aqueous solution, we need to consider the dissociation of aluminum chloride (AlCl3) in water.

The balanced chemical equation for the dissociation of aluminum chloride is:

AlCl3(aq) → Al3+(aq) + 3 Cl-(aq)

From the equation, we can see that for each formula unit of aluminum chloride that dissociates, it will produce one Al3+ ion and three Cl- ions.

To find the number of moles of aluminum chloride in the solution, we can use the formula:

moles = concentration (M) × volume (L)

Given:
Concentration (M) = 0.490 M
Volume (L) = 113 mL = 0.113 L

moles of AlCl3 = 0.490 M × 0.113 L = 0.05537 moles

Since each formula unit of aluminum chloride produces one Al3+ ion, the number of Al3+ ions in the solution is also 0.05537 moles.

Similarly, each formula unit of aluminum chloride produces three Cl- ions. So, the number of Cl- ions in the solution is 3 times the number of moles of AlCl3:

moles of Cl- = 3 × moles of AlCl3 = 3 × 0.05537 moles = 0.16611 moles

To convert the moles of ions to the number of ions, we can use Avogadro's number, which is 6.022 × 10^23 ions/mole.

Number of Al3+ ions = moles of Al3+ × Avogadro's number
= 0.05537 moles × 6.022 × 10^23 ions/mole
= 3.334 × 10^22 ions

Number of Cl- ions = moles of Cl- × Avogadro's number
= 0.16611 moles × 6.022 × 10^23 ions/mole
= 9.993 × 10^22 ions

Therefore, in the given aqueous solution of 113 mL of 0.490 M aluminum chloride, there are approximately 3.334 × 10^22 Al3+ ions and 9.993 × 10^22 Cl- ions.