Some Basic Concepts of Chemistry - Result Question 2
####3. In which case is number of molecules of water maximum?
[2018]
(a) $18 mL$ of water
(b) $0.18 g$ of water
(c) $10^{-3} mol$ of water
(d) $0.00224 L$ of water vapours at $1 atm$ and $273 K$
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Solution:
- (a)
$\therefore 30$ moles of $H_2$ are required.
(1) Mass of water $=18 \times 1=18 g$
Molecules of water $=$ mole $\times N_A$
$=\frac{18}{18} N_A=N_A$
(2) Molecules of water $=$ mole $\times N_A$
$=\frac{0.18}{18} N_A=10^{-2} N_A$
$\begin{cases} 2 \\ \text{ 管 } \end{cases} .$
