Question

Volume occupied by one molecule of water is: (Density=\[1gmc{{m}^{-3}}\])
(A) \[3\times {{10}^{-23}}c{{m}^{3}}\]
(B) \[5.5\times {{10}^{-23}}c{{m}^{3}}\]
(C) \[9\times {{10}^{-23}}c{{m}^{3}}\]
(D) \[6.023\times {{10}^{-23}}c{{m}^{3}}\]

Answer 1

Hint: Density relates mass and volume of the molecule. If we know any two of the quantities out of these two, we can find the third unknown quantity. Molecular weight of water is 18\[gmmo{{l}^{-1}}\].

Complete step by step solution:
We are given that density of water is \[1gmc{{m}^{-3}}\] so that 1 gm of water occupies 1 \[c{{m}^{3}}\] of volume.
So, we can say that \[Density=\dfrac{Mass}{volume}\]
To find volume occupied by one molecule of water, we will need to find the absolute weight of one molecule of water. Let’s find it.
We know that the molecular weight of water is 18\[gmmo{{l}^{-1}}\] means that 1 mole of water has a mass of 18 gm.
As 1 mole involves \[6.023\times {{10}^{23}}\] numbers of molecules, we can write that
If weight of \[6.023\times {{10}^{23}}\] molecules of water is 18gm,
Then weight of 1 molecule of water will be\[=\dfrac{1\times 18}{6.023\times {{10}^{23}}}\]
Weight of 1 molecule of water\[=2.99\times {{10}^{-23}}gm\]
Now put that mass of 1 molecule of water into the formula of density.
\[Density=\dfrac{Mass}{volume}\]
\[1gmc{{m}^{-3}}=\dfrac{2.99\times {{10}^{-23}}gm}{volume}\]
Volume\[=\dfrac{2.99\times {{10}^{-23}}gm}{1gmc{{m}^{3}}}\]
Volume=\[2.99\times {{10}^{-23}}c{{m}^{3}}\]
Our answer is closest to \[3\times {{10}^{-23}}c{{m}^{3}}\].

So, correct answer is (A) \[3\times {{10}^{-23}}c{{m}^{3}}\]

Additional information:
There is another short way to solve this question.
First, we will find the mass of 1 molecule of water and then as density is equal to \[1gmc{{m}^{-3}}\], we can say that the mass of one molecule of water in gm will be equal to the volume occupied by one molecule of water expressed in \[c{{m}^{3}}\] unit. These both are equal because density is \[1gmc{{m}^{-3}}\].

Note: Note that the weight of one molecule of water is not 18 gm but it is the weight of one mole of water molecules. This is often confusing. Make sure that you remember Avogadro's number rightly, which is \[6.023\times {{10}^{23}}\].