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How many molecules of water and oxygen atoms are present in $0.9g$ of water?A.$3.010 \times {10^{ - 20}}$B.$3.010 \times {10^{22}}$C.$3.010 \times {10^{20}}$D.$3.010 \times {10^{ - 22}}$

Last updated date: 16th Apr 2024
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Answer
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Hint: At first we should know about the number of molecules present in one atom, which is nothing but the Avogadro number of molecules present. The Avogadro number is $6.023 \times {10^{23}}$.

Complete answer:
At first, we should know about the number of molecules present in one atom, which is nothing but the Avogadro number of molecules present. The Avogadro number is $6.023 \times {10^{23}}$.
Complete step by step answer:-In the question they gave the amount of water in which we should find out the number of molecules of water and oxygen atoms.
First, we should calculate the number of moles of water, where the number of moles (n) is given by given weight by the actual weight of water.
Therefore,
Moles of water $= \dfrac{{0.9}}{{18}} = 0.05$
In one mole of water, the Avogadro number of water molecules are present, so for $0.05$moles of water
Number of water molecules $= 0.05 \times 6.023 \times {10^{23}} = 3.010 \times {10^{22}}$
The one water molecule contains one oxygen and two hydrogen atoms.
In the question, they asked to calculate for one oxygen atom so already we calculated for one water molecule which is also equal to one oxygen atom.
Then the number of molecules of water and oxygen atoms present in $0.9g$ of water is $3.010 \times {10^{22}}$.

So the answer is B.

Note:
In order to calculate these types of questions, you should always keep in mind that one mole of an atom contains an Avogadro number of particles and an Avogadro number of molecules.