In which case the number of molecules of water maximum?

In which case the number of molecules of water maximum?
  1. 0.00224 L of water vapours at 1 atm and 273 K.
  2. 0.18 g of water.
  3. 18 mL of water.
  4. 10-3 mol of water.

Solution

(i) 18 mL water:
As  dH2O=1g/mL So   WH2O=18g
nH2O=1818=1
Molecules =1×NA
(ii) 0.18 g of water:
nH2O=0.1818=0.01
H2O molecules=0.01×NA
(iii)  (VH2O (g))STP=0.00224 L
nH2O=V22.4=0.0022422.4=0.0001
Molecules =0.0001×NA
(iv)   nH2O=10-3
H2O molecules=10-3×NA

Asked in: NEET 2018

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