Question

The volume of water molecule is (Take, density of water is ${{10}^{3}}$ kg${{m}^{-3}}$ and Avogadro’s number = $6\times {{10}^{23}}mol{{e}^{-1}}$).
A. $3\times {{10}^{-28}}{{m}^{3}}$
B. $3\times {{10}^{-29}}{{m}^{3}}$
C. $1.5\times {{10}^{-28}}{{m}^{3}}$
D. $1.5\times {{10}^{-29}}{{m}^{3}}$

Answer 1

Hint: First we will find the mass of water in a unit volume. After that is done we will calculate the mass of one mole of water. Then we will find out how many moles of water are there in that mass and finally, we will find the number of molecules in that volume using Avogadro’s number. Then we will find the volume of one molecule of water.

Complete step-by-step solution:
First, let us calculate the mass of water in one unit volume i.e. in one ${{m}^{3}}$. We have the density of water as ${{10}^{3}}$ kg${{m}^{-3}}$. So, we easily get the mass of water in one ${{m}^{3}}$of volume as 1000 kilograms. Now we will find the mass of one mole of water.
The water molecule is made up of 2 atoms of Hydrogen and one atom of oxygen. Their atomic masses add up to 18 and that is how much one mole of water will weigh in grams.
In 1000 kilograms of water there will be $\dfrac{1000\times 1000}{18}=55555.56$ moles of water.
Multiplying this number by the Avogadro’s number we will get the total number of water molecules present in one ${{m}^{3}}$ of water.
total number of water molecules present in one ${{m}^{3}}$=$55555.56\times 6\times {{10}^{23}}=3.33\times {{10}^{28}}$
So, the volume of one molecule of water will be equal to $\dfrac{1}{3.33\times {{10}^{28}}}=3\times {{10}^{-29}}{{m}^{3}}$

Hence, the correct option is B, i.e., $3\times {{10}^{-29}}{{m}^{3}}$.

Note: We are dealing with big numbers here so it is better to write them in scientific notification when possible. Take care to avoid calculation mistakes. The whole process must be done in order to avoid confusion and calculation mistakes.