Question

In which case is the number of molecules of water maximum?
A. ${10^{ - 3}}$ mol of water.
B. 0.00224 L of water vapour at 1 atm and 273 k.
C. 0.18 g of water.
D. 18 ml of water.

Answer 1

Hint: To solve this question, we have to remember that the number of molecules can be calculated by the formula, number of molecules = number of moles $ \times $ ${N_A}$, where ${N_A}$ = Avogadro number = $6.022 \times {10^{23}}$ and the number of moles can be calculated by using the formula, number of moles = $\dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
First, we have to find out the molar mass of water.
So, molecular formula of water = ${H_2}O$
Molecular mass = $2\left( {{\text{molar mass of hydrogen}}} \right) + {\text{molar mass of oxygen}}$
Molecular mass = $2\left( 1 \right) + 16$
Molecular mass = 18 g.
Therefore, 1 mole of water = 18 g of water.
Let us find the number of molecules of water for each option.

(A). ${10^{ - 3}}$ mol of water.
Given that, Number of moles = ${10^{ - 3}}$
Number of molecules = number of moles $ \times $$6.022 \times {10^{23}}$,
Putting the value, we will get
Number of molecules = ${10^{ - 3}} \times 6.022 \times {10^{23}}$
Number of molecules = $6.022 \times {10^{20}}$

(B). 0.00224 L of water vapour at 1 atm and 273 k.
We know that, according to the ideal gas equation
$ \Rightarrow n = \dfrac{{pV}}{{RT}}$, where n is the number of moles, p is the pressure, V is the volume, T is the temperature and R is the gas constant, R = $0.0821$L atm ${K^{ - 1}}mo{l^{ - 1}}$
Here we have, V = 0.00224 L, p = 1 atm and T = 273 K
Putting all these values in above formula, we will get
$ \Rightarrow n = \dfrac{{1 \times 0.00224}}{{0.0821 \times 273}}$
$ \Rightarrow n = {10^{ - 4}}$moles.
So,
Number of molecules = ${10^{ - 4}} \times 6.022 \times {10^{23}}$
Number of molecules = \[6.022 \times {10^{19}}\]

(C). 0.18 g of water.
We know that,
Number of moles = $\dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$
Putting the values,
Number of moles = $\dfrac{{0.18}}{{18}} = {10^{ - 2}}$moles.
Then,
Number of molecules = ${10^{ - 2}} \times 6.022 \times {10^{23}}$
Number of molecules = \[6.022 \times {10^{21}}\]

(D). 18 ml of water.
18 ml of water = 18 g of water. [ mass = volume$ \times $density, common density of water = 1 gm/ml ]
Number of moles = 1 mole
So, Number of molecules = $1 \times 6.022 \times {10^{23}}$
Number of molecules = $6.022 \times {10^{23}}$
Hence, from all the values, we can see that the number of molecules of water is maximum $6.022 \times {10^{23}}$ in 18ml of water.
So, the correct answer is “Option D”.

Note: Whenever we ask such a question, we should know some basic concepts of chemistry. In case of gases, a mole is defined as the amount of the gas which occupies a volume of 22.4 L at STP, which is called its molar volume and the mass of one mole of substance in gram is called its molar mass.