What is the total number of ions present in a 9.5\text{ g} sample of pure magnesium chlori… — JAMB UTME

What is the total number of ions present in a $9.5\text{ g}$ sample of pure magnesium chloride ( $\text{MgCl}_2$ )? $[\text{Mg} = 24, \text{Cl} = 35.5, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

$1.81 \times 10^{23}$ Answer
B
$6.02 \times 10^{22}$
C
$1.20 \times 10^{23}$
D
$2.24 \times 10^{23}$

Answer

1.81 \times 10^{23}

ions

The sample contains

0.1\text{ mol}

\text{MgCl}_2

. Because each formula unit of

\text{MgCl}_2

dissociates into 3 ions (

1\text{ Mg}^{2+}

and

2\text{ Cl}^-

), there are

0.3\text{ mol}

of total ions. Multiplying

0.3\text{ mol}

by Avogadro's constant (

6.02 \times 10^{23}

) yields

1.81 \times 10^{23}

ions.

Step-by-Step Solution

Calculate the molar mass of magnesium chloride (

\text{MgCl}_2

)

Molar mass

= 24 + 2(35.5) = 95\text{ g/mol}

Molar mass is required to convert the given mass into moles.

Determine the number of moles of

\text{MgCl}_2

9.5\text{ g}

Moles

= 9.5\text{ g} / 95\text{ g/mol} = 0.1\text{ mol}

Applying the mole formula:

\text{moles} = \text{mass} / \text{molar mass}

Determine the total moles of ions released upon dissociation

\text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^-

, giving 3 moles of ions per mole of salt. Total ion moles

= 0.1 \times 3 = 0.3\text{ mol}

Each formula unit of magnesium chloride contains one magnesium ion and two chloride ions.

Calculate the total number of individual ions using Avogadro's constant

Total ions

= 0.3 \times 6.02 \times 10^{23} = 1.806 \times 10^{23} \approx 1.81 \times 10^{23}

ions

Multiplying the total moles of ions by Avogadro's constant yields particle count.

Key Concept

The Mole Concept and Avogadro's Constant applied to Ionic Dissociation

Estimated Time:1m 30s

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