How many moles and numbers of ions of each type are present in the following aqueous solutions? (a) 88 mL of 1.75 M magnesium chloride,
(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L,
(c) 1.65 L of a solution containing 8.83×10218.83×1021 formula units of cesium nitrate per liter.

a. 0.154 moles = 2.78x10²³ ions

b. 2.18x10⁻⁴ moles = 6.56x10²⁰ ions

c. 1.77x10²² ions = 0.0147 moles

Explanation:

The multiplication of volume of solution in liters times molarity gives the moles of solute.

a. Moles MgCl₂:

88mL = 0.088L * (1.75mol / L) = 0.154 moles MgCl₂

Uaing Avogadro's number:

0.154mol  *(6.022x10²³ molecules / 1 mol) = 9.27x10²² molecules.

As 1 molecule of MgCl₂ contains 3 ions:

Ions = 2.78x10²³

b. Moles Al₂(SO₄)₃:

Aluminium sulfate has a molar mass of 324.15g/mol:

0.321L * (0.22g Al₂(SO₄)₃ / L) * (1mol / 324.15g) = 2.18x10⁻⁴ moles.

1 molecule contains 5 ions:

2.18x10⁻⁴ moles  *(6.022x10²³ molecules / 1 mol) * 5 = 6.56x10²⁰ ions.

c. 8.83x10²¹ formula units of CsNO₃ = Number of molecules. Ions are:

8.83x10²¹ * 2 = 1.77x10²² ions

And moles are:

8.83x10²¹ Formula units * (1mol /  6.022x10²³) = 0.0147 moles

A) is the correct one:)