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How can I calculate the number of hydrogen atoms in one gallon deionized water?

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Last updated date: 29th Feb 2024
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Hint: Water from which all of its ions are removed is called deionized water. Deionized water does not possess any charge. Deionized water has many applications in industry. Gallon is a unit of volume similar to litre.

Complete step by step answer:
For solving the question, first let us convert the gallon to litre. One gallon is equal to $3.785$ litre. Hence one gallon of water is equal to $3.785$ L of water.
Now we need to find how many grams of water is present in one litre. Density of water is $1000g/L$. Hence we can write,
Density $ = \dfrac{{mass}}{{volume}}$
Or,
Mass $ = $ density $ \times $ volume
Mass $ = 1000 \times 3.785 = 3785g$
That is, $3785$ grams of water is present in one litre. Now we need to find the number of moles of water present. The equation is,
$n = \dfrac{w}{m}$
Where n is the number of moles, w is the weight and M is molar mass.
Molar mass of water is $18g/mol$ .
Hence,
$n = \dfrac{{3785}}{{18}} = 210.28$ moles
That is $210.28$ moles of water is present. One mole of every substance contains $6.022 \times {10^{23}}$ number molecules. Hence $210.28$ moles of water will contain $210.28 \times 6.022 \times {10^{23}}$ number of water molecules.
That is, Number of water molecules present $ = 210.28 \times 6.022 \times {10^{23}} = 1.266 \times {10^{26}}$ .
Each water molecule contains two hydrogen atoms. Hence $1.266 \times {10^{26}}$ molecules of water will contain $2 \times 1.266 \times {10^{26}}$ hydrogen atoms.
That is, the number of hydrogen atoms $ = 2 \times 1.266 \times {10^{26}} = 2.53 \times {10^{26}}$ .

Hence the number of hydrogen atoms in one gallon deionized water is $2.53 \times {10^{26}}$ .

Note: Hydrogen exists as a diatomic molecule. Hence we can say that, the number of hydrogen molecules required to form one gallon of deionized water will be $1.266 \times {10^{26}}$ .

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