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) \[\text{1}{{\text{0}}^{\text{--3}}}\text{ mol}\]of water
Correct Answer: C
Solution :
| [c] [a] Moles of water\[\text{=}\frac{0.00224}{22.4}={{10}^{-4}}\] |
| Molecules of water \[\text{=mole }\!\!\times\!\!\text{ }{{\text{N}}_{\text{A}}}\text{=1}{{\text{0}}^{\text{-4}}}{{\text{N}}_{\text{A}}}\] |
| [c] Molecules of water\[\text{=mole }\!\!\times\!\!\text{ }{{\text{N}}_{\text{A}}}\text{=}\frac{0.18}{18}{{\text{N}}_{\text{A}}}\] |
| \[\text{=1}{{\text{0}}^{\text{-2}}}{{\text{N}}_{\text{A}}}\] |
| [c] \[\text{Mass of water=18 }\!\!\times\!\!\text{ 1=18 g}\] |
| Molecules of water \[\text{=mole }\!\!\times\!\!\text{ }{{\text{N}}_{\text{A}}}\text{=}\frac{\text{18}}{\text{18}}{{\text{N}}_{\text{A}}}\] |
| \[\text{=}{{\text{N}}_{\text{A}}}\] |
| [d] Molecules of water \[\text{=mole }\!\!\times\!\!\text{ }{{\text{N}}_{\text{A}}}\text{=1}{{\text{0}}^{\text{-3}}}{{\text{N}}_{\text{A}}}\] |
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