Question

The density of ${H_2}O$ is 1$gm{l^{ - 1}}$. What is the volume occupied by 1 molecule of water.
\[{\text{1 mol }} \to {\text{18g = 6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}\]

Answer 1

Hint:The density of a substance is the amount of that substance occupied per unit volume. To solve this problem, we need to know about some basic concepts like one mole contains $6.022 \times {10^{23}}$ molecules.

Complete step by step answer:
It is given that the density of water is 1 $gm{l^{ - 1}}$ which means one gram of water molecules occupy one mL of volume. We know that the molar mass of ${H_2}O$ is 18 g.
We know that 18 g of water contains $6.022 \times {10^{23}}$ molecules of water.
So, 1 g of water contains $\dfrac{{6.022 \times {{10}^{23}}}}{{18}} = 0.334 \times {10^{23}}$ molecules of ${H_2}O$. That means $0.334 \times {10^{23}}$ molecules of ${H_2}O$ will occupy a volume of 1 ml. From this, we can find the volume occupied by 1 molecule of${H_2}O$. That is, $\dfrac{1}{{0.334 \times {{10}^{23}}}} = 2.99 \times {10^{ - 23}}$ ml.
Therefore, the volume occupied by 1 molecule of ${H_2}O$ is $2.99 \times {10^{23}}$ ml.


Note:
A mole is defined as the amount of substance containing the same number of discrete entities (atoms, molecules, ions, etc.) .The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol).