Question

Calculate the number of Magnesium ions present in 0.06 g of Magnesium Sulphate?

Answer 1

2.99 × 10²⁰ ≅  3 × 10²⁰ Mg²⁺ ions

Explanation:

1 mole of MgSO₄ = 120.36 g

Number of moles = 0.06/120.36

                               = 4.98 × 10⁻⁴ moles

1 mole contains 6.022 × 10²³ atoms

4.98 × 10⁻⁴ will contain = 4.98 × 10⁻⁴ × 6.022 × 10²³

                                        = 2.99 × 10²⁰

The number of Mg²⁺ present in 1 molecule of MgSO₄ is 1

Hence the number of Mg²⁺ will be 1 × 2.99 × 10²⁰ ≅  3 × 10²⁰

Answer 2

Answer:

The number of Magnesium ions present in 0.06 g of Magnesium Sulfate is 2.99 × $10^{20}$

Explanation:

The chemical formula of magnesium sulfate is $Mg SO_{4}$

The molar mass of magnesium sulfate = 120.366 g/mol

The given mass of magnesium sulfate = g

∴ The number of moles in 0.06 g of magnesium sulfate = $\frac{0.06 }{120.366}$

                                    = 4.98 × $10^{-4}$ moles

one mole of magnesium sulfate contains 6.022 × $10^{23}$ molecules

∴ 4.98 × $10^{-4}$ moles contain 2.99 × $10^{20}$ molecules

The number of magnesium ions present in magnesium sulfate = 1

number of Magnesium ions present in 0.06 g of Magnesium Sulfate

                    = 1 × 2.99 × $10^{20}$ = 2.99 × $10^{20}$ ions