A) \[9.0\times {{10}^{-}}^{23}c{{m}^{3}}\]
B) \[6.023\times {{10}^{-}}^{23}c{{m}^{3}}\]
C) \[3.0\times {{10}^{-}}^{23}c{{m}^{3}}\]
D) \[5.5\times {{10}^{-}}^{23}c{{m}^{3}}\]
Correct Answer: C
Solution :
| \[6.02\times {{10}^{23}}\] molecules of water = 1 mol |
| = 18 g |
| \[\therefore \] Mass of one molecule of water |
| \[=\frac{18}{6.023\times {{10}^{23}}g}\] |
| \[\because \] \[d=\frac{m}{V}\] |
| \[\therefore \] \[V=\frac{m}{d}=\frac{18}{6.023\times {{10}^{23}}\times 1}\] |
| \[\approx 3\times {{10}^{-23}}c{{m}^{3}}\] |
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