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Question: How many water molecules are present in one ml of water (d = 1 gm/ml) ? (A) \(5.55 \times {10^{21}...

How many water molecules are present in one ml of water (d = 1 gm/ml) ?
(A) 5.55×10215.55 \times {10^{21}}
(B) 6.023×10236.023 \times {10^{23}}
(C) 3.346×10223.346 \times {10^{22}}
(D) 6.023×10226.023 \times {10^{22}}

Explanation

Solution

Hint: First of all calculate the weight of water in 1 ml of water and its moles. Since 1 mole of any substance contains 6.023×10236.023 \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=mvd = \dfrac{m}{v}
1=m11 = \dfrac{m}{1} ,
m=m = 11
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=118n = \dfrac{1}{{18}} moles
We can thus say that for 1 ml of water the number of moles is 118\dfrac{1}{{18}}.
-We all know that: in 1 mole of any substance the number of molecules present is = NA{N_A}
= 6.023×10236.023 \times {10^{23}}
So, now we will use the unitary method to calculate the number of molecules present in 118\dfrac{1}{{18}} moles of water.
In 1 mole of water → 6.023×10236.023 \times {10^{23}} molecules are present
So, in 118\dfrac{1}{{18}} moles of water → 6.023×1023×1186.023 \times {10^{23}} \times \dfrac{1}{{18}} molecules
=0.3346×1023= 0.3346 \times {10^{23}} molecules
=3.346×1022= 3.346 \times {10^{22}} molecules
So, we can conclude that in 1 ml of water the number of molecules present is 3.346×10223.346 \times {10^{22}}.
Hence the correct option is: (C) 3.346×10223.346 \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 6.022×10236.022 \times {10^{23}} (NA{N_A} - Avogadro number). The expression of moles will be:
n=Wt.givenMol.wt.n = \dfrac{{Wt.given}}{{Mol.wt.}}
=Vol.givenVol.(STP)= \dfrac{{Vol.given}}{{Vol.(STP)}}
=No.NA= \dfrac{{No.}}{{{N_A}}}