Question

If the total mass of oxygen present in CuSO4.5H2O is 160 g, then calculate the total number of electrons present in CuSO4.5H2O​

Answer 1

Answer:

• 4.8895 × 10²⁵ electrons.

Explanation:

Given :

Total mass of oxygen present in CuSO₄.5H₂O is 160 g.

To find :

Total number of electrons present in CuSO₄.5H₂O.

Solution :

We know Atomic masses of,

⇒ Cu = 63.5 g

⇒ S = 32 g

⇒ O = 16 g

⇒ H = 1 g

Finding molar mass,

⇒ 3.5 + 32 + 4(16) + 5 (2)(1) + (16)

⇒ 63.5 + 32 + 64 + 90

⇒ 249.5 g

Finding the number of moles,

⇒ n = Given Mass/Molar Mass

⇒ n = 160 g/249.5 g

⇒ n = 0.64 moles

We know,

1 mole = 6.023 × 10²³ molecules

∵ Avogadro number of molecules.

⇒ 0.64 moles = ?

⇒ 0.64 moles = 0.64 ×6.023 × 10²³

⇒ 0.64 moles = 3.85 × 10²³ molecules

We know,

Number of electrons = Atomic number

⇒ Cu = 1 × 29 = 29 electrons  

⇒ S = 1 × 16 = 16 electrons  

⇒ O = 9 × 8 = 72 electrons  

⇒ H = 10 × 1 = 10 electrons

∴ Total number of electrons are 127 electrons

Now,

1 molecule = 127 electrons

⇒ 3.85 × 10²³ molecules = ?

⇒ 3.85 × 10²³ molecules = 3.85 × 10²³ molecules × 127

⇒ 3.85 × 10²³ molecules = 4.8895 × 10²⁵ electrons

∴ Number of electrons are 4.8895 × 10²⁵ electrons.

Answer 2

Answer:

Given Substance: CuSO₄. 5 H₂O

Given Mass: 160 g

Atomic Mass of constituting elements are:

Cu = 63.5 g

S = 32 g

O = 16 g

H = 1 g

Hence Molar Mass of CuSO₄. 5 H₂O is:

⇒ 63.5 + 32 + 4(16) + 5 ( 2(1) + 16 )

⇒ 63.5 + 32 + 64 + 5 ( 18 )

⇒ 249.5 g

Calculating the number of moles of the given compound, we get:

⇒ n = Given Mass / Molar Mass

⇒ n = 160 g / 249.5 g

⇒ n = 0.64 moles

Now let us calculate the number of molecules in 0.64 moles of CuSO₄.5H₂O. We know that, 1 mole of any compound has Avogadro number of molecules. Hence 0.64 moles would have:

⇒ 1 mole : Nₐ :: 0.64 mole = ?

⇒ x = 0.64 × Nₐ

⇒ 0.64 × 6.023 × 10²³

⇒ 3.85 × 10²³ molecules

Let us calculate the number of electrons present in 1 molecule of CuSO₄.5H₂O. We know that, number of electrons present in an element is equal to it's atomic number. Hence we get:

⇒ Cu = 1 × 29 = 29 electrons

⇒ S = 1 × 16 = 16 electrons

⇒ O = 9 × 8 = 72 electrons

⇒ H = 10 × 1 = 10 electrons

Total number of electrons in 1 molecule of CuSO₄.5H₂O is:

⇒ (29 + 16 + 72 + 10) = 127 electrons.

Therefore, if 1 molecule contains 127 electrons, then 3.85 × 10²³ molecules would contain:

$\implies x = \dfrac{127 \times 3.85 \times 10^{23}}{1}\\\\\\\implies \boxed{ \bf{ x = 4.8895 \times 10^{25} \: \text{electrons}}}$

This is the required answer.