Question

In which case is number of molecules of water maximum?

• 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 1

The correct option is C 18 mL of water

(1) moles of water = $\frac{0.00224}{22.4}$ = ${10}^{-4}$
Molecules of water = $mole\times N_A$ = ${10}^{-4}$$N_A$

(2) Molecules of water = $mole\times N_A$ = $\frac{0.18}{18}$$N_A$ = ${10}^{-2}$$N_A$

(3) Mass of water = $18\times 1$ = 18 g
Molecules of water = $mole\times N_A$ = $\frac{18}{18}$$N_A$ = $N_A$

(4) Molecules of water = $mole\times N_A$ = ${10}^{-3}$$N_A$