Question

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

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

Answer 1

Answer: (B)

1 mole of water $=6.022\times 10^{23}$ molecules $=18$ gram of water (1) $10^{-3}mol$ of water $=6.022\times 10^{20}$ molecules
(2) $V=0.00224L,P=1atm,T=273K$
$\begin{array}{rl}\therefore n=\frac{PV}{RT}& =\frac{1\times 0.00224}{0.0821\times 273}=10^{-4}\text{ moles }\\ & =6.022\times 10^{19}\text{ molecules }\end{array}$
(3) $0.18g$ of water $=10^{-2}$ moles
$=6.022\times 10^{21}\text{ molecules }$
(4) $18ml$ of water $=18g$ of water $\because$ density of water $=1\frac{g}{ml}$
$\begin{array}{l}=1\text{ mole }\\ =6.022\times 10^{23}\text{ molecules }\end{array}$
$\therefore 18ml$ of water has the maximum number of water molecules.