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# 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) $6.023 \times {10^{22}}$

Last updated date: 11th Sep 2024
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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.
-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: $d = \dfrac{m}{v}$
$1 = \dfrac{m}{1}$ ,
$m =$ $1$
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: $n = \dfrac{1}{{18}}$ moles
We can thus say that for 1 ml of water the number of moles is $\dfrac{1}{{18}}$.
-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 $\dfrac{1}{{18}}$ moles of water.
In 1 mole of water → $6.023 \times {10^{23}}$ molecules are present
So, in $\dfrac{1}{{18}}$ moles of water → $6.023 \times {10^{23}} \times \dfrac{1}{{18}}$ 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:
$n = \dfrac{{Wt.given}}{{Mol.wt.}}$
$= \dfrac{{Vol.given}}{{Vol.(STP)}}$
$= \dfrac{{No.}}{{{N_A}}}$