Question

Assuming the density of water to be $1\,g/ml,$ calculate the volume occupied one molecule of water.
(A) $2.98 \times {10^{ - 23}}\,c{m^3}$
(B) $2.99 \times {10^{ - 22}}\,c{m^3}$
(C) $1.98 \times {10^{ - 23}}\,c{m^3}$
(D) $4.98 \times {10^{ - 23}}\,c{m^3}$

Answer 1

Hint: The smallest unit into which a substance can be divided into, without changing its chemical composition is called “molecule”. Use the concept of molar mass to calculate the number of molecules in $1\,g$ of water. And then, use it to calculate the volume occupied by one molecule of water.

Complete Step by Step Solution:
The density of water is approximately $1\,g/ml.$
Thus we can say that, $1\,g$ of water occupied $1\,ml$ volume.
We know that, the molar mass of water is $ = 18\,g/mol.$
Therefore, the number of moles of water in $1\,g = \dfrac{1}{{18}}$ moles
$ = 0.055$ moles ${H_2}O$
Since, molar mass is the number of moles present in a particular mass of a compound. And since, molar mass of water is $18\,g/mole$. We can say that $18\,g$ of water contains $6.022 \times {10^{23}}\,moles$ of water.
Where,
$6.022 \times {10^{23}}$ is known as Avogadro’s number.
Thus, $1g$ water will have$ = \dfrac{{6.022 \times {{10}^{23}}}}{{18}}$ moles of water.
$ = 0.334 \times {10^{23}}$ moles of water.
That means, $0.334 \times {10^{23}}$ molecules of water will occupy $1\,ml$ of volume.
Therefore, volume occupied by 1 water molecule is
$ = \dfrac{1}{{0.334 \times {{10}^{23}}}}$
$ = 2.99 \times {10^{ - 23}}\,ml$
Thus volume occupied by 1 water molecule is $2.99 \times {10^{ - 23}}\,ml$

Therefore, from the above explanation the correct option is (A) $2.98 \times {10^{ - 23}}\, c{m^3}.$

Note:
To solve this question, you need to know how to calculate molar mass. You cannot solve it if you don’t know the molar mass of water. Also, Avogadro’s number is a constant, which says that the number of molecules present in the molar mass of any compound are the same. $1ml = 1c{m^3}$. Therefore, our answer didn’t change even though we got it in terms of $ml$.