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Q.
In which case the number of water molecules is maximum?
NTA AbhyasNTA Abhyas 2022
Solution:
Number of moleules $ \propto $ Number of moles
(A) Moles of water $=\frac{0.00224}{22.4}=10^{- 4}$
Molecules of water $=mole\times N_{A}=10^{- 4}N_{A}$
(B) Molecules of water $=mole\times N_{A}=\frac{0.18}{18}N_{A}=10^{- 2}N_{A}$
(C) Mass of water $=18\times 1=18g$
Molecules of water $=mole\times N_{A}=\frac{18}{18}N_{A}=N_{A}$
(D) Molecules of water $= 10^{- 2} \text{mole} = 10^{- 2} \text{N}_{\text{A}}$
Hence 18 ml $H_{2}O$ has maximum molecules.