NEET Solved Paper 2018 Question 33
Question: In which case is number of molecules of water maximum? [NEET - 2018]
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) $ 1{0^{\text{–3}}}\text{ mol} $ of water
Show Answer
Answer:
Correct Answer: C
Solution:
[a] Moles of water $ \text{=}\frac{0.00224}{22.4}={10^{-4}} $
Molecules of water $ \text{=mole }\times{N_A}\text{=1}{0^{\text{-4}}}{N_A} $
[b] Molecules of water $ \text{=mole }\times{N_A}\text{=}\frac{0.18}{18}{N_A} $
$ \text{=1}{0^{\text{-2}}}{N_A} $
[c] $ \text{Mass of water=18 }\times\text{ 1=18 g} $
Molecules of water $ \text{=mole }\times{N_A}\text{=}\frac{18}{18}{N_A} $ $ \text{=}{N_A} $
[d] Molecules of water $ \text{=mole }\times{N_A}\text{=1}{0^{\text{-3}}}{N_A} $
