Question

In which case the number of water molecules is 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. $10^{-2}$ mol of water

Answer 1

Answer: (C)

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.