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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.

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: $d = \dfrac{m}{v}$

$1 = \dfrac{m}{1}$ ,

$m =$ $1$

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:

$n = \dfrac{{Wt.given}}{{Mol.wt.}}$

$= \dfrac{{Vol.given}}{{Vol.(STP)}}$

$= \dfrac{{No.}}{{{N_A}}}$

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: $d = \dfrac{m}{v}$

$1 = \dfrac{m}{1}$ ,

$m =$ $1$

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:

$n = \dfrac{{Wt.given}}{{Mol.wt.}}$

$= \dfrac{{Vol.given}}{{Vol.(STP)}}$

$= \dfrac{{No.}}{{{N_A}}}$

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