Question

Number of molecules in $ 1{\text{L}} $ of water is close to
(A) $ \dfrac{{18}}{{22.4}} \times {10^{23}} $
(B) $ 55.5 \times 6.023 \times {10^{23}} $
(C) $ \dfrac{{6.023}}{{23.4}} \times {10^{23}} $
(D) $ 18 \times 6.023 \times {10^{23}} $

Answer 1

Hint : To solve this question, we have to find out the total number of moles in the given volume of water by using the value of the density of water. Then, using Avogadro's constant we can find out the required number of molecules of water.

Formula Used: The formula which is used in solving this question is given by
 $\Rightarrow n = \dfrac{m}{{{M_0}}} $ , here $ n $ is the number of moles of a substance whose mass is $ m $ and has a molar mass of $ {M_0} $ .

Complete step by step answer
We know that the density of water at STP is equal to $ 1g/c{m^3} $ . So we have
 $\Rightarrow d = 1g/c{m^3} $
Now, we also know that one cubic centimetre volume is equal to one millilitre volume, that is,
 $\Rightarrow 1c{m^3} = 1{\text{ml}} $
So the density of water is
 $\Rightarrow d = 1g/{\text{ml}} $
So this means that one millilitre volume of water contains one gram of water.
Now we know that
 $\Rightarrow 1{\text{ml}} = 1{{\text{0}}^{ - 3}}{\text{L}} $
So $ 1{{\text{0}}^{ - 3}}{\text{L}} $ of water contains one gram of water.
This means that mass of $ 1{\text{L}} $ water is given by
 $\Rightarrow m = \dfrac{1}{{{{10}^{ - 3}}}}g $
 $ \Rightarrow m = 1000g $
Now, as we know that the number of moles of a substance is given by the formula
 $\Rightarrow n = \dfrac{m}{{{M_0}}} $
From (i) the mass of water is equal to $ 1000g $ . Also the molar mass of water is equal to $ 18g/{\text{mol}} $ . Making these substitutions in the above formula, we get
 $ n = \dfrac{{1000}}{{18}} $
 $ \Rightarrow n = 55.55{\text{ mol}} $
From Avogadro’s law, we know that in one mole of a substance, there are Avogadro numbers of molecules of that substance present. So, the total number of molecules of water in $ 55.55 $ moles of water is given by
 $\Rightarrow N = 55.5{{\text{N}}_{\text{A}}} $
We know that the Avogadro’s number is $ {{\text{N}}_{\text{A}}} = 6.023 \times {10^{23}} $ . So we get
 $\Rightarrow N = 55.5 \times 6.023 \times {10^{23}} $
Thus, these are the required number of molecules $ 1{\text{L}} $ of water.
Hence, the correct answer is option B.

Note
There was no information given in the question regarding the temperature and pressure of the water. So we ourselves assumed the condition of standard temperature and pressure and therefore we were able to use the standard value of density of water.