Question

Calculate the total number of ions present in 50.5g of NaCl.
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Answer 1

Answer :-

1.035 × 10²⁴ ions are there in 50.5 g of NaCl.

Explanation :-

We have :-

→ Mass of NaCl = 50.5 g

Also :-

→ Molar mass of Sodium = 23 g/mol

→ Molar mass of Chlorine = 35.5 g/mol

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Molar mass of NaCl is :-

= 23 + 35.5

= 58.5 g/mol

Let's calculate the number of moles in 50.5 g of NaCl :-

= Given Mass/Molar mass

= 50.5/58.5

= 505/585

= 0.86 mole

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Now, we know that a molecule is NaCl has 1 ion of Na⁺ and 1 ion of Cl⁻. So, 1 molecule of NaCl has 2 ions.

So total moles of ions :-

= 2 × 0.86

= 1.72 moles of ions .

Required number of ions :-

= Moles of ions × Avogadro Number

= 1.72 × 6.022 × 10²³

= 10.35 × 10²³

= 1.035 × 10²⁴ ions

Answer 2

Answer:

1.0398 × 10²⁴ ions

Explanation:

Given ,

50.5g of NaCl

To Find :-

Total number of Ions persent in 50.5g of NaCl

How To Do :-

As we know how to calculate molecular weight of 'NaCl' and after calculating we know that for every compund if normal Molecular weight there will 6.023 × 10²³ ions . So we need to calculate how much of ions will be present in 50.5g of NaCl.

Solution :-

Molecular weight of NaCl = Atomic weight of 'Na' + Atomic weight of 'Cl'

= 23 + 35.5

= 58.5

So , for 58.5g there will be 6.023 × 10²³ ions.

We need to find for 50.5 g , Let it be 'x'

58.5g - 6.023 × 10²³

50.5g - x

Cross Multiplication :-

58.5g(x) = 6.023×10²³(50.5g)

58.5g(x) = 304.1615g × 10²³

x = 304.1615g × 10²³/58.5g

x = 5.199 × 10²³ ions

Multiplying with '2' because there are 2 ions in NaCl. [ Number of Ions = Total Number of Ions × Avegadro number]

= 2 × 5.199 × 10²³ ions

= 10.398 × 10²³ ions

= 1.0398 × 10²⁴ ions

There will be 1.0398 × 10²⁴ ions present in the 50.5g of NaCl.