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Question: In which case, the number of molecules of water is maximum: A ) 0.00224 L of water vapours at 1 at...

In which case, the number of molecules of water is maximum:
A ) 0.00224 L of water vapours at 1 atm and 273 K.
B ) 18 mL of water.
C ) 10−3 mol{10^{ - 3}}{\text{ mol}} of water.
D ) 0.18 g of water.

Explanation

Solution

One mole of water contains Avogadro’s number of molecules which is 6.023×1023 molecules6.023 \times {10^{23}}{\text{ molecules}}. The weight of one mole of water is equal to its molar mass, which is 18 g/mol. At 1 atm and 273 K, one mole of any gas occupies a volume of 22.4 L.

Complete answer:
A) At 1 atm and 273 K, one mole of any gas occupies a volume of 22.4 L.

At 1 atm and 273 K, a volume of 0.00224 L corresponds to 0.00224L22.4L/mol=10−4moles\dfrac{{0.00224{\text{L}}}}{{22.4{\text{L/mol}}}} = {10^{ - 4}}{\text{moles}} of water.
10−4moles{10^{ - 4}}{\text{moles}} of water will contain 6.023×1023 molecules/mol× 10−4 mol1 mol = 6.023×1019 molecules6.023 \times {10^{23}}{\text{ molecules/mol}} \times {\text{ }}\dfrac{{{{10}^{ - 4}}{\text{ mol}}}}{{1{\text{ mol}}}}{\text{ = }}6.023 \times {10^{19}}{\text{ molecules}}.
of water vapours at 1 atm and 273 K{\text{1 atm and 273 K}}.

B) 18 mL of water is around 18 g of water because the density of water is around 1 g/mL.
The molar mass of water is 18 g/moL. Thus, 18 g of water contains one mole. One mole of water contains Avogadro’s number of molecules which is 6.023×1023 molecules6.023 \times {10^{23}}{\text{ molecules}}.

C) One mole of water contains Avogadro’s number of molecules which is 6.023×1023 molecules6.023 \times {10^{23}}{\text{ molecules}}. 10−3 mol{10^{ - 3}}{\text{ mol}} of water will contain 6.023×1023 molecules/mol× 10−3 mol1 mol = 6.023×1020 molecules6.023 \times {10^{23}}{\text{ molecules/mol}} \times {\text{ }}\dfrac{{{{10}^{ - 3}}{\text{ mol}}}}{{1{\text{ mol}}}}{\text{ = }}6.023 \times {10^{20}}{\text{ molecules}}.

D) The molar mass of water is 18 g/mol. Thus, 18 g of water contains one mole. One mole of water contains Avogadro’s number of molecules which is 6.023×1023 molecules6.023 \times {10^{23}}{\text{ molecules}}. 0.18 g of water corresponds to 0.18 g0.18 g/mol=10−2 mol\dfrac{{0.18{\text{ g}}}}{{0.18{\text{ g/mol}}}} = {10^{ - 2}}{\text{ mol}}.
of water. One mole of water contains Avogadro’s number of molecules which is 6.023×1023 molecules6.023 \times {10^{23}}{\text{ molecules}}. 10−2 mol{10^{ - 2}}{\text{ mol}} of water will contain 6.023×1023 molecules/mol× 10−2 mol1 mol = 6.023×1021 molecules6.023 \times {10^{23}}{\text{ molecules/mol}} \times {\text{ }}\dfrac{{{{10}^{ - 2}}{\text{ mol}}}}{{1{\text{ mol}}}}{\text{ = }}6.023 \times {10^{21}}{\text{ molecules}}.
The number of water molecules is maximum for 18 mL18{\text{ mL}} of water.

Hence, the option B ) 18 mL of water represents the correct answer.

Note: Either mass of water or volume of water or number of moles of water can be converted to the number of molecules of water. If the volume of water vapour at a particular temperature and pressure is given, then it can also be converted to the number of molecules of water.