Question

Density of water at room temperature is 1.0 $gc{m^{ - 3}}$. How many molecules are there in one drop of water if its volume is 0.1 $c{m^3}$?

Answer 1

Hint: The number of molecules in a drop of water can be calculated by the formula-
number of molecules = number of moles $ \times $ \[6.022 \times {10^{23}}\]molecules
But for this, we need to know the number of moles present which will be calculated as-
number of moles = $\dfrac{{Mass{\text{ of one drop}}}}{{Molar{\text{ mass of the water}}}}$

Complete step by step answer:
First, let us start by writing about the density of a molecule. What density is and on what factors it depends.
The density of a molecule can be defined as the mass of the substance present per unit volume. It is denoted by the symbol $\rho $. The mathematical equation for density can be written as -
Density of a molecule = $\dfrac{{Mass{\text{ of the substance}}}}{{Volume{\text{ of the substance}}}}$

We are given :
Density of water at room temperature = 1.0 $gc{m^{ - 3}}$
We have to calculate:
The number of molecules present in one drop of water
And the volume of drop = 0.1 $c{m^3}$
Thus, we need to calculate the number of molecules present in 0.1 $c{m^3}$ of water.
For calculating the number of molecules, one should first know about the number of moles present. So, we will first calculate the number of moles of water present in 0.1 $c{m^3}$.
We have, number of moles = $\dfrac{{Mass{\text{ of one drop}}}}{{Molar{\text{ mass of the water}}}}$
The mass of one drop of water can be calculated from density of water as-

We have, Density of a molecule = $\dfrac{{Mass{\text{ of the substance}}}}{{Volume{\text{ of the substance}}}}$
Thus, Mass of one drop of water = Density of water $ \times $ Volume of one drop of water
Mass of one drop of water = 1.0 $gc{m^{ - 3}}$$ \times $ 0.1 $c{m^3}$
Mass of one drop of water = 0.1 g
Now, we can calculate the number of moles = $\dfrac{{Mass{\text{ of one drop}}}}{{Molar{\text{ mass of the water}}}}$

Thus, putting all the values.
We have, number of moles = $\dfrac{{0.1g}}{{18g}} = 0.00555$
We know, number of molecules = number of moles $ \times $ \[6.022 \times {10^{23}}\]molecules
number of molecules = 0.00555 $ \times $ \[6.022 \times {10^{23}}\]molecules
number of molecules = \[0.0334 \times {10^{23}}\]molecules
Thus, the one drop of water will contain $0.0334 \times {10^{23}}$ molecules.

Note: The term $6.022 \times {10^{23}}$ is the Avogadro number. It can be represented by ${N_A}$. It is assumed that one mole of any substance will contain $6.022 \times {10^{23}}$ number of molecules or ions or atoms and further this value will be equal to 22.4 L for gases at STP (Standard Temperature Pressure).