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How many unit cells are present in a cube-shaped ideal crystal ofNaCI of mass 1 .00 g? [Atomic masses : Na = 23, Cl = 35.5] |
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Answer» `5.14xx10^(21)` unit CELLS `=(1.0)/(58.5)xx6.02xx10^(23)` since in NACL TYPE of structure. 4 formula units form 'a' cell `therefore `unit cells`=(1.0xx6.02xx10^(23))/(58.5xx4)=2.57xx10^(21)` unit cells. |
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