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

Options

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^(-3)` mol of water

Answer

C

Explanation

Number of molecules = Mole `xx` Avogadro's number `(N_(A))`
The number of molecules of water in each of the given options is calculated as
(i) 18 mL of water
Number of moles `(n_(H_(2)O))`
`=("Mass of substance in g"(W_(H_(2)O)))/("Molar mass in g mol"^(-1)(M_(H_(2)O)))`
`W_(H_(2)O)=18g`
`[because " Density of water"(d_(H_(2)O))=1 g L^(-1)]`
`therefore n_(H_(2)O)=(18)/(18)=1`
Number of molecules of water `=1xxN_(A)`
(ii) 0.18 g of water
`n_(H_(2)O)=(W_(H_(2)O))/(M_(H_(2)O))=(0.18)/(18)=0.01`
Number of molecules of water `=0.01xxN_(A)`
(iii) 0.00224 L of water vapours at 1 atm and 273 K. At STP [1 atm and 273 K],
Number of moles [with reference to volume]
`=("Volume of gas in litres")/(22.4)`
`=(0.00224)/(22.4)=0.0001`
Number of molecules of water `=0.0001xxN_(A)`
`(iv) 10^(-3)` mol of water
Number of molecules of water `=10^(-3)xxN_(A)`
`therefore` Among the given options, option(i) contains the maximum number of water molecules.