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In which case is number of molecules of water maximum?

0.00224 L of water vapours at 1 atm and 273 K
0.18 g of water
18 mL of water
$10^{–3}$ mol of water
Explanation:
(1) Moles of water =$\dfrac{0.00224}{22.4}=10^{-4}$
Molecules of water = $mole \times N_{A}=10^{-4} N_A$
(2) Molecules of water = $mole × N_A =\dfrac{0.18}{18}N_A$
$10^{-2}N_A$
(3) Mass of water = 18 × 1 = 18 g
Molecules of water = mole × $N_A =\dfrac{18}{18}N_A$
$=N_A$
(4) Molecules of water = mole × $N_A = 10^{-3}N_A$
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