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

(1) Moles of water `=(0.00224)/(22.4)=10^(-4)`
Molecules of water =mole `xx N_(A)=10^(-4)N_(A)`
(2) Molecules of water =mole `xxN_(A)=(0.18)/(18)N_(A)`
`=10^(-2)N_(A)`
(3) Mass of water `=18xx1=18` g
Molecules of water =mole `xx N_(A)=(18)/(18)N_(A)`
`=N_(A)`
(4) Molecules of water =mole `xxN_(A)=10^(-13)N_(A)`