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In which case is number of molecules of water maximum?

A

0.00224 L of water vapours at 1 atm and 273 K

B

0.18 g of water

C

18 mL of water

D

10−310^{-3} mol of water

Step-by-Step Solution

(1) Moles = 0.0022422.4=10−4\frac{0.00224}{22.4} = 10^{-4}, Molecules = 10−4NA10^{-4} N_A. (2) Moles = 0.1818=10−2\frac{0.18}{18} = 10^{-2}, Molecules = 10−2NA10^{-2} N_A. (3) Mass = 18×1=18 g18 \times 1 = 18 \text{ g}, Moles = 1818=1\frac{18}{18} = 1, Molecules = NAN_A. (4) Moles = 10−310^{-3}, Molecules = 10−3NA10^{-3} N_A. Option (3) has the maximum number of molecules.

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