A) \[2.57\times {{10}^{21}}\]
B) \[5.14\times {{10}^{21}}\]
C) \[1.28\times {{10}^{21}}\]
D) \[1.71\times {{10}^{21}}\]
Correct Answer: A
Solution :
(i) NaCI contains FCC unit cell therefore the unit cell of the crystal will be FCC of unit cell. Find the mass by the formula mass = volume \[\times \] density. (ii) Calculate the number of unit cell presents in one gram of crystal. Mass of one unit-cell (m) = volume \[\times \] density \[={{a}^{3}}\times d={{a}^{3}}\times \frac{Mz}{{{N}_{0}}{{a}^{3}}}=\frac{Mz}{{{N}_{0}}}\] \[m=\frac{58.5\times 4}{6.02\times {{10}^{23}}}g\] \[\therefore \] Number of unit cells in 1 g \[=\frac{1}{m}\] \[=\frac{6.02\times {{10}^{23}}}{58.5\times 4}=2.57\times {{10}^{21}}\]
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