Positive and negative charge in 18 cc of water is how much?
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Positive and negative charge in 18 cc of water is how much?
Your answer
Given
- amount of water = 18 cc = 18 mL
- ( as 1 cc = 1 mL )
Focusing on the given molecule ,
i.e Water ( H2O )
We know that ,
- Total no. of electrons present in water = 10
- i. e. 2 hydrogen atom contains = 2 electrons
- and 1 oxygen atom contains = 8 electrons
So there are 10 electrons.
Now,
For No. of moles of water of water ,
- Weight / Mass of the given water = Density x volume
- As the density of the water = 1 g/mL and the volume of the water is = 18 mL
- So Weight = 1 x 18 = 18 grams
- Molecular weight of water (H2O) = 18 g/mol
So , now we can easily calculate the moles of water as we have both the original weight and molecular weight of water.
- No. of moles of water = Weight of water / Molecular weight of water
- = 18 / 18 = 1 mol
As the water have 10 electrons ,
- so no. of moles of electrons in 1 mole of water molecule = 10 x 1 = 10
Now , 1 mol of water contains electrons
- = no. of moles of electrons x Avogadro's number
- = 10 x 6.023 x 10²³
- = 6.023 x 10²⁴
So we have 6.023 x 10²⁴ electrons in 1 mol of water.
Now , we can easily calculate the charge on 1 mole of water or 6.023 x 10²⁴ electrons by multiply it with the official charge on an electron.
i.e
- Total charge on electrons in given water molecule = electron's no. x charge on an electron
- = 6.023 x 10²⁴ x ( - 1.6 x 10^-19 ) C
- = - 9.632 x 10^5 C
And same for the total charge involves protons
- Because water is a stable molecule and the no. of electrons and protons are equal also the charge on an electron and proton is equal So ,
- Charge ( Protons ) = 9.632 x 10^5 C
So the total / net Charge on water molecule
- = 9.632 x 10^5 C + ( -9.632 x 10^5 C )
- = 0 (zero)
- i. e stability / neutrality of Water / H2O molecule
And for this question
- The positive charge present = 9.632 x 10^5 C
- The negative charge present = -9.632 x 10 ^5 C
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