What is the actual volume occupied by water molecules present in 20 $c{m^3}$ of
water?
(A) $20c{m^3}$
(B) $40c{m^3}$
(C) $10c{m^3}$
(D) $24.89d{m^3}$

Hint: A water molecule is formed by bonding between two hydrogen and one oxygen atoms. Water molecules are attached by hydrogen bonding. A hydrogen atom in one water molecule is attached to the lone pair of electrons on an oxygen atom of a neighbouring water molecule. Due to hydrogen bonding, half the volume occupied by water molecules is actually unoccupied.

Complete step by step solution:
Here, we need to calculate the actual volume that is occupied by water molecules present in 20$c{m^3}$ of water. We can calculate this by using density of water, volume of water and the number of moles of water present in 20$c{m^3}$.
The density of water is the measure of its mass per unit volume. It is denoted by $\rho $.The density of water is $1g/c{m^3}$.
The quantity of liquid present at definite temperature and pressure is measured in terms of its volume. It is denoted as V.
Let us first write down the given quantities.
$\rho = 1g/c{m^3}$
$V = 20c{m^3}$
Step 1: To calculate the number of moles of water molecules, we need molar mass and mass of water.
To calculate mass of water, we will use the formula of density.
$\rho = \dfrac{m}{V}$
On rearranging the above formula, we get,
$m = \rho \times V$
On substituting the respective values,
$m = 1g/c{m^3} \times 20c{m^3}$
$m = 20g$
Therefore, the $20c{m^3}$ of water molecules weighs 20g.
Step 2: Now, we have got the mass of water molecules, we need to calculate the molar mass of water.
The molar mass of water is equal to that of molecular weight of water.
Molecular weight of water $ = (2 \times 1) + (1 \times 16)$
Molecular weight of water $ = 18$
$\therefore M = 18$
The molar mass of water is 18.
Step 3: Now, we have got respective values to calculate the number of moles of water. The amount of substance present in a sample is defined by the term ‘mole’.
Let us first write the formula for calculating the number of moles of water in a given sample.
$n = \dfrac{m}{M}$
Putting the respective values in the above formula, we get,
$n = \dfrac{{20}}{{18}}$
$\therefore n = 1.111moles$
Therefore, the number of moles of water in $20c{m^3}$ water is 1.111moles.
As we know, the volume occupied by 1 mole of a substance is $22.4L$.
To find out the volume occupied by 1.11moles of water, we need to multiply it with 22.4L.
The volume occupied by 1.111moles of water $ = 22.4 \times 1.111$
The volume occupied by 1.111moles of water $ = 24.89L$
Also, $1L = 1d{m^3}$
Therefore,
The volume occupied by 1.111moles of water $ = 24.89d{m^3}$
Therefore, the actual volume occupied by water molecules present in $20c{m^3}$ of water is $24.89d{m^3}$.

Hence, the correct answer is option (D).

Note: Generally, on cooling the liquids contracts, reducing their volume. But, this is not true in case of water. When water is cooled, it will contract till it reaches 4°C. After 4°C, when the temperature of water drops further, water will expand slightly till it reaches its freezing point that is 0°C. On freezing, it will expand further.