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

Options

A. `10^(-3)` mol of water
B. 18 mL of water
C. 0.00224 L of water vapours at 1 atm and 273K
D. 0.18g of water

Answer

B

Explanation

(1) molecular of water
=number of moles`xxN_(A)=10^(-3)N_(A)`
[`N_(A)`=Avogadro number]
(2) Mass of water=volume`xx`density=18`xx`1g=18g
`therefore`molecularr of water=number of moles `xxN_(A)=(18)/(18)xxN_(A)=N_(A)`
(3) Number of moles of water`=(0.00224)/(22.4)=10^(-4)`
`therefore`Molecular of water=number of moles`xxN_(A)=10^(-4)N_(A)`
(4) Molecules of water=number of moles`xxN_(A)=(0.18)/(18)N_(A)=10^(-2)N_(A)`