Calculate the number of sodium ions present in 0.031 g of sodium oxide.
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Given:-Mass of sodium oxide =0.031g
Molar mass of sodium oxide(Na2O) :-62 g Number of sodium ions present in 0.031 g of sodium oxide =0.031/62*6.022*10^23 =6.022*10^20 Na+ ions
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Given:
The mass of sodium oxide (Na₂O) = 0.031g
To Find:
The number of sodium ions present
Solution:
We know that the number of moles of a substance = Given mass / molar mass
The molar mass of Na₂O = 2 X The molar mass of Na + 1 X The molar mass of O
= 2 X 23 + 16
= 62g
So the number of moles of Na₂O = 0.031 / 62
=0.0005
From the formula, we can infer that 1 mole of Na₂O will have 2 moles of Na cations and one mole of oxygen anions.
⇒ 0.0005 moles of Na₂O will have 2 X 0.0005 moles of Na⁺ ions
= 0.001
Now, 1 mole of any substance has 6.022 X 10²³ constituents
⇒ 0.001 moles of Na⁺ ions will have 6.022 X 10²³ X 0.001 ions
= 6.022 X 10²⁰ ions
Thus, the number of sodium ions present in 0.031 g of sodium oxide is 6.022 X 10²⁰.
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