Question

How many water molecules are present in one ml of water (d = 1 gm/ml) ?
(A) $5.55\times {10}^{21}$
(B) $6.023\times {10}^{23}$
(C) $3.346\times {10}^{22}$
(D) $6.023\times {10}^{22}$

Answer 1

Hint: First of all calculate the weight of water in 1 ml of water and its moles. Since 1 mole of any substance contains $6.023\times {10}^{23}$ atoms/molecules; use the unitary method to calculate the number of water molecules in the obtained number of moles of water.
Complete step by step answer:
-First of all, we will calculate the total number of moles of water we are talking about.
According to the question the given volume of water = 1 ml; and the density of water is 1 gm/ml. We know that: $d={\displaystyle \frac{m}{v}}$
$1={\displaystyle \frac{m}{1}}$ ,
$m=$ $1$
From this we can say that the weight of 1 ml of water will be 1 gm.
The molecular weight of water is = 18 gm/mol
The number of moles of water will be: $n={\displaystyle \frac{1}{18}}$ moles
We can thus say that for 1 ml of water the number of moles is ${\displaystyle \frac{1}{18}}$.
-We all know that: in 1 mole of any substance the number of molecules present is = $N_A$
= $6.023\times {10}^{23}$
So, now we will use the unitary method to calculate the number of molecules present in ${\displaystyle \frac{1}{18}}$ moles of water.
In 1 mole of water → $6.023\times {10}^{23}$ molecules are present
So, in ${\displaystyle \frac{1}{18}}$ moles of water → $6.023\times {10}^{23}\times {\displaystyle \frac{1}{18}}$ molecules
$=0.3346\times {10}^{23}$ molecules
$=3.346\times {10}^{22}$ molecules
So, we can conclude that in 1 ml of water the number of molecules present is $3.346\times {10}^{22}$.
Hence the correct option is: (C) $3.346\times {10}^{22}$
Note: At these STP conditions the volume of 1 mole of gas will always be 22.4 litre and the number of atoms or molecules in it will be $6.022\times {10}^{23}$ ($N_A$ - Avogadro number). The expression of moles will be:
$n={\displaystyle \frac{Wt.given}{Mol.wt.}}$
$={\displaystyle \frac{Vol.given}{Vol.(STP)}}$
$={\displaystyle \frac{No.}{N_A}}$