A) \[1.5\times {{10}^{23}}\]
B) \[1.5\times {{10}^{22}}\]
C) \[3.0\times {{10}^{22}}\]
D) \[3.0\times {{10}^{23}}\]
Correct Answer: B
Solution :
Density of unit cell \[(d)=\frac{Z\times M}{{{a}^{3}}.{{N}_{A}}}\] Mass of unit cell,\[(m)=\]volume\[\times \]density \[={{a}^{3}}\times \frac{Z\times M}{{{a}^{3}}.{{N}_{A}}}\] \[=\frac{Z\times M}{{{N}_{A}}}=\frac{4\times 58.5}{6.023\times {{10}^{23}}}\] \[=3.885\times {{10}^{-22}}\] Therefore, number of unit cells in\[5.85\text{ }g\text{ }NaCl\]crystals \[=\frac{5.85}{3.885\times {{10}^{-22}}}=1.5\times {{10}^{22}}\]
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