Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How many electrons are present in the \[18\]ml of water?

seo-qna
Last updated date: 22nd Jul 2024
Total views: 349.8k
Views today: 5.49k
Answer
VerifiedVerified
349.8k+ views
Hint: We have to remember that the water is one of the important things for surveys in the world. Without water nothing in the world. Plant photosynthesis also depends on the water. In Human being metabolism, water plays a vital role. The molecular formula of water is \[{{\text{H}}_{\text{2}}}{\text{O}}\].
Formula used:
We have to know that the density of the material depends on the mass and volume of the material. Density of the material is equal to the mass of the material divided by the volume of the material.
\[{\text{density = }}\dfrac{{{\text{mass}}}}{{{\text{volume}}}}\]
Moles are defined as the given mass of the molecule is divided by the molecular mass of the molecule.
$Moles = \dfrac{{{\text{Mass of the molecule}}}}{{{\text{Molecular weight of the molecule}}}}$
The molecular weight of the molecule is dependent on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atom in the molecule.
${\text{Molecular weight}} = {\text{Number of atoms}} \times {\text{Atomic weight of the atom}}$
The number of molecules is equal to the number of moles of the molecules multiplied by Avogadro’s number. The numerical value of Avogadro’s number is \[6.022 \times {10^{23}}\].
\[{\text{The Number Of Molecules = number of moles \times 6}}{\text{.022 \times 1}}{{\text{0}}^{{\text{23}}}}\]

Complete answer:
The given volume of water is \[18{\text{ml}}\]
The density of water is \[{\text{1gm}}{{\text{l}}^{{\text{ - 1}}}}\]
Calculate the mass of the given water is
\[{\text{density = }}\dfrac{{{\text{mass}}}}{{{\text{volume}}}}\]
We change the formula for our concern
\[{\text{mass = density}} \times {\text{volume}}\]
Now we can substitute the known values we get,
\[ = 18 \times 1\]
On simplification we get,
\[ = 18\]
The mass of the given water is \[18g\]
Calculate the molecular weight of water,
Water has one oxygen and two hydrogen. The atomic weight of oxygen is \[16g\]. The atomic weight of hydrogen is \[1g\].
${\text{Molecular weight}} = {\text{Number of atoms}} \times {\text{Atomic weight of the atom}}$
Now we can substitute the known values we get,
\[ = 16 \times 1 + 1 \times 2\]
On simplification we get,
\[ = 18\]
The molecular weight of the water is \[18g\]
Calculate the moles of the given water,
\[moles\, = \dfrac{{mass\,of\,the\,molecule}}{{molecular\,weight\,of\,the\,molecule}}\]
Now we can substitute the known values we get,
\[ = \dfrac{{18}}{{18}}\]
On simplification we get,
\[ = 1\]
One moles of water present in a given volume.
The calculate the number of molecules of water in the given volume,
\[{\text{The Number Of Molecules = number of moles \times 6}}{\text{.022 \times 1}}{{\text{0}}^{{\text{23}}}}\]
Now we can substitute the known values we get,
\[ = 1 \times 6.022 \times {10^{23}}\]
On simplification we get,
$ = 6.022 \times {10^{23}}$
The given volume of water contains \[6.022 \times {10^{23}}\] molecules.
Each molecule of water consists of ten electrons. Because oxygen having eight electrons and hydrogen has one electron and water having two hydrogen. Hence entirely water having ten electrons in one molecule. The molecular formula of water is\[{{\text{H}}_{\text{2}}}{\text{O}}\].
Now we can substitute the known values we get,
\[ = 10 \times 6.022 \times {10^{23}}\]
On simplification we get,
\[ = 6.022 \times {10^{24}}\]
Hence, \[10ml\] of water contains \[6.022 \times {10^{24}}\] electrons.

Note:
We have to know that the moles are the one of the main units in chemistry. The moles of the molecule depend on the mass of the molecule and molecular mass of the molecule. Chemical reactions are measured by moles only. The number of equivalents of the reactant also depend on the moles of the molecule. The number of moles of the reactant and product are equal in the equilibrium reaction.