A) \[0.0432\text{ }kg\text{ }mo{{l}^{-1}}\]
B) \[0.016\text{ }kg\text{ }mo{{l}^{-1}}\]
C) \[0.4320\text{ }kg\text{ }mo{{l}^{-1}}\]
D) \[0.0305\text{ }kg\text{ }mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[\rho =\frac{Z\times M}{{{a}^{3}}\times {{N}_{A}}}\] \[9\times {{10}^{3}}=\frac{4\times M}{{{(200\sqrt{2}\times {{10}^{-12}})}^{3}}\times 6\times {{10}^{23}}}\] \[M=\frac{9\times {{10}^{3}}\times 16\times \sqrt{2}\times 6\times {{10}^{-7}}}{4}\]\[=0.0305kg/mol\]
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