Question

How many moles and numbers of ions of each type are present in the following aqueous solution? 63.1 mL of 1.85 M magnesium chloride:

Answer 1

We have 0.117 moles Mg^2+ and 0.234 moles Cl-

The number of Mg^2+ ions = 7.04 *10^22 ions

The number of Cl- ions = 1.4*10^23 ions

Step-by-step explanation:

Step 1: Data given

Volume of magnesium chloride = 63.1 mL

Concentration of solution = 1.85 M

Step 2: Calculate moles MgCl2

Moles MgCl2 = concentration * volume

Moles MgCl2 = 1.85 M * 0.0631 L

Moles MgCl2 = 0.117 moles

Step 3: Calculate moles of ions

MgCl2 → Mg^2+ + 2Cl-

For 1 mol MgCl2 we have 1 mol Mg^2+ and 2 moles Cl-

For 0.117 moles MgCl2 we have 0.117 moles Mg^2+ and 2*0.117 = 0.234 moles Cl-

The number of Mg^2+ ions = 0.117 * 6.022 *10^23 = 7.04 *10^22 ions

The number of Cl- ions = 1.4*10^23 ions