Question

Calculate no. of chloride ions in 9.5g of magnesium chloride.

Answer 1

(9.5 g MgCl2) * (1 mole MgCl2 / 95.211 g MgCl2) * 
(6.022 x 10^23 MgCl2 molecules / 1 mole MgCl2) * (2 Cl ions / 1 molecule MgCl2) 

= 1.2 x 10^23 chloride ions

Answer 2

9.5 g of Magnesium chloride contains $1.201\times10^{23}$ chloride ions.

Given:

Mass of Magnesium chloride = 9.5 g

To find:

Number of chloride ions = ?

Formula to be used:

• Moles = $\frac{Given mass}{Molecular weight}$
• Number of ions = Moles × Avogadro's number

Calculation:

This sum must be solved in three steps

1. Number of moles calculation:

Moles = $\frac{Given mass}{Molecular weight}$

Moles = $\frac{9.5}{95.2} = 0.09$

2. Total number of ions calculation:

Number of ions = Moles × Avogadro's number

Number of ions = 0.09 × $6.022\times10^{23}$

Number of ions = $0.600\times10^{23}$

3. Total number of chloride ion calculation:

As there are 2 $Cl^-$ ions present in magnesium chloride,

No. of chloride ions = $2\times0.600\times10^{23}$

No. of chloride ions = $1.201\times10^{23}$

Conclusion:

The number of chloride ions in 9.5g of magnesium chloride is calculated as $1.201\times10^{23}$.

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