Question

In which case, the number of molecules of water maximum?

• A. $18\text{mL}$ of water
• B. $0.18\text{mL}$ of water
• C. ${10}^{-3}\text{mol}$ of water
• D. $0.00224\text{L}$ of water vapours at $1\text{atm}$ and $273\text{K}$

Answer 1

The correct option is A $18\text{mL}$ of water

$\left(a\right)$ Mass of water $=d\times V$$=1\times 18$$=18\text{g}$
Moles of water $=\frac{w_{H_{2}O}}{M_{H_{2}O}}$ $=\frac{18}{18}=$ 1 mol
Molecules of water $=\text{mole}\times N_A$ $=1\times N_A$$=N_A$

$\left(b\right)$ Molecules of water $=\text{mole}\times N_A$ $=\frac{0.18}{18}{N}_A$
$={10}^{-2}{N}_A$

$\left(c\right)$ Molecules of water $=\text{mole}\times N_A$ $={10}^{-3}{N}_A$

$\left(d\right)$ Moles of water $=\frac{0.00224}{22.4}$$={10}^{-4}\text{mol}$
Molecules of water $=\text{mole}\times N_A$ $={10}^{-3}{N}_A$

Hence, the correct answer is option (a).