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Question: In which case is the number of molecules of water maximum? A. \({10^{ - 3}}{\text{mol}}\) of water...

In which case is the number of molecules of water maximum?
A. 10−3mol{10^{ - 3}}{\text{mol}} of water
B. 0.00224L0.00224{\text{L}} of water vapor at 1atm1{\text{atm}} and 273K273{\text{K}}
C. 0.18g0.18{\text{g}} of water
D. 0.18mL0.18{\text{mL}} of water

Explanation

Solution

Stoichiometry is the study of the quantitative aspects of chemical reactions. Chemical equations are concise representations of chemical reactions. Mole is defined as the quantity of a substance that contains the same number of ultimate particles as are present in 12g12{\text{g}} of carbon−12 - 12.

Complete step by step answer:
Concentration is the amount of solute dissolved in a given amount of solution. There are different types of concentration units. Formula and molecular mass deal with individual atoms and molecules. Mole is the unit that relates the number of particles and mass.
The molar volume of water in gaseous form is 22.4L22.4{\text{L}}.
Therefore one mole of water is 18g18{\text{g}} contains 22.4L22.4{\text{L}} volume and 6.022×10236.022 \times {10^{23}} molecules.
A. 1mol1{\text{mol}} of water has 6.022×10236.022 \times {10^{23}} molecules.
For 10−3mol{10^{ - 3}}{\text{mol}} of water, the number of molecules will be 10−3×(6.022×1023)=6.022×1020{10^{ - 3}} \times \left( {6.022 \times {{10}^{23}}} \right) = 6.022 \times {10^{20}}.
B. 22.4L22.4{\text{L}} volume will have 6.022×10236.022 \times {10^{23}} molecules.
Therefore 0.00224L0.00224{\text{L}} volume will have 6.022×102322.4L×0.00224L = 1.35×102122.4=6.022×1019\dfrac{{6.022 \times {{10}^{23}}}}{{22.4{\text{L}}}} \times 0.00224{\text{L = }}\dfrac{{1.35 \times {{10}^{21}}}}{{22.4}} = 6.022 \times {10^{19}} number of molecules.
C. 18g18{\text{g}} of water contains 6.022×10236.022 \times {10^{23}} molecules.
So 0.18g0.18{\text{g}} of water contains 6.022×102318g×0.18g = 1.08×1023g18g=6.022×1021\dfrac{{6.022 \times {{10}^{23}}}}{{18{\text{g}}}} \times 0.18{\text{g = }}\dfrac{{1.08 \times {{10}^{23}}{\text{g}}}}{{18{\text{g}}}} = 6.022 \times {10^{21}} number of molecules.
D. 22.4L22.4{\text{L}} volume will have 6.022×10236.022 \times {10^{23}} molecules.
Therefore 0.18mL0.18{\text{mL}} of water contains 6.022×102322.4L×0.18L = 1.08×102322.4=4.8×1021\dfrac{{6.022 \times {{10}^{23}}}}{{22.4{\text{L}}}} \times 0.18{\text{L = }}\dfrac{{1.08 \times {{10}^{23}}}}{{22.4}} = 4.8 \times {10^{21}} number of molecules.
Among these options, more value is for 0.18g0.18{\text{g}} of water which contains 6.022×10216.022 \times {10^{21}} number of molecules.

So, the correct answer is Option C.

Note:
Mole concepts enable us to solve stoichiometric problems involving mass relations of reactants and products in chemical reactions. Moles provide a bridge from molecular scale to real-world scale. Water molecule has the same molar mass when it is in solid, liquid or gaseous state. So it is applicable for all the molar conceptual values.