
Volume occupied by one molecule of water (density 1 gm $cm^-3$) is
A.)$3\times {{10}^{-23}}$ $cm^3$
B.)$5.5\times {{10}^{-23}}$$cm^3$
C.)$9\times {{10}^{-23}}$ $cm^3$
D.)$6.023\times {{10}^{-23}}$ $cm^3$
Answer
607.5k+ views
Hint: 1 mole of molecules has the same mass as the gram molecular mass of the substance. 1 mole = $6.023\times {{10}^{23}}$ number of particles.
Complete step by step answer:
We know that the gram molecular mass of water or $H_2O$ is 18 grams. This contains 1 mole of molecules, that is, $6.023\times {{10}^{23}}$molecules.
Therefore, the mass of $6.023\times {{10}^{23}}$molecules of water is 18 grams.
Therefore, mass of 1 molecule of water in grams will be
$\dfrac{18}{6.023\times {{10}^{23}}}=2.99\times {{10}^{-23}}gm$
Now, the given density of water is 1 g $cm^-3$.
Now, we know that $Volume=\dfrac{Mass}{Density}$
Therefore, volume of one molecule of water will be
$\dfrac{Mass\text{ }of\text{ }one\text{ }molecule\text{ }of\text{ }water}{density\text{ }of\text{ }water}$
$\dfrac{2.99\times {{10}^{-23}}}{1}=2.99\times {{10}^{23}}c{{m}^{3}}$
Therefore the volume of one molecule of water is $2.99\times {{10}^{-23}}$ $cm^3$.
The closest option to this value is the correct option A) $3\times {{10}^{-23}}$ $cm^3$
Additional information:
$6.023\times {{10}^{23}}$is called the Avogadro’s number (NA). One mole of any substance, be it molecules or any other particles contains Avogadro number of particles. Avogadro number of molecules (or atoms) of all chemical elements and compounds have the mass of their gram molecular weight.
Note: Students must be extra careful while solving these types of questions with the Avogadro number involved. Many times writing the number wrong can lead to silly mistakes. For example, a classic silly mistake is writing the Avogadro number as $6.023\times {{10}^{-23}}$ instead $6.023\times {{10}^{23}}$. This can lead to a completely different and absurd answer.
Students must also understand the various relations in chemical compounds, especially gases of the Avogadro number (i.e. 1 mole) with their gram molecular mass, atomic mass and volume (in case of gases). The mole also comes in handy while balancing chemical reactions.
Complete step by step answer:
We know that the gram molecular mass of water or $H_2O$ is 18 grams. This contains 1 mole of molecules, that is, $6.023\times {{10}^{23}}$molecules.
Therefore, the mass of $6.023\times {{10}^{23}}$molecules of water is 18 grams.
Therefore, mass of 1 molecule of water in grams will be
$\dfrac{18}{6.023\times {{10}^{23}}}=2.99\times {{10}^{-23}}gm$
Now, the given density of water is 1 g $cm^-3$.
Now, we know that $Volume=\dfrac{Mass}{Density}$
Therefore, volume of one molecule of water will be
$\dfrac{Mass\text{ }of\text{ }one\text{ }molecule\text{ }of\text{ }water}{density\text{ }of\text{ }water}$
$\dfrac{2.99\times {{10}^{-23}}}{1}=2.99\times {{10}^{23}}c{{m}^{3}}$
Therefore the volume of one molecule of water is $2.99\times {{10}^{-23}}$ $cm^3$.
The closest option to this value is the correct option A) $3\times {{10}^{-23}}$ $cm^3$
Additional information:
$6.023\times {{10}^{23}}$is called the Avogadro’s number (NA). One mole of any substance, be it molecules or any other particles contains Avogadro number of particles. Avogadro number of molecules (or atoms) of all chemical elements and compounds have the mass of their gram molecular weight.
Note: Students must be extra careful while solving these types of questions with the Avogadro number involved. Many times writing the number wrong can lead to silly mistakes. For example, a classic silly mistake is writing the Avogadro number as $6.023\times {{10}^{-23}}$ instead $6.023\times {{10}^{23}}$. This can lead to a completely different and absurd answer.
Students must also understand the various relations in chemical compounds, especially gases of the Avogadro number (i.e. 1 mole) with their gram molecular mass, atomic mass and volume (in case of gases). The mole also comes in handy while balancing chemical reactions.
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