A) \[3.506 gc{{m}^{-\,3}}\]
B) \[3.0 gc{{m}^{-\,3}}\]
C) \[1.75 gc{{m}^{-\,3}}\]
D) \[1.5 gc{{m}^{-\,3}}\]
Correct Answer: A
Solution :
We know that \[\rho =\frac{ZM}{N{{a}^{3}}}\] Given, \[\operatorname{Z}=8,\,\,M\,\,=12\] \[\operatorname{N} =6.023 \times 1{{0}^{23}}\] \[\operatorname{a}=\,\,3.569\,\,\times \,\,1{{0}^{-\,10}}\,m\] \[\rho =\frac{8\,\,\times \,\,12}{6.023\,\,\times \,\,{{10}^{23}}\,{{(3.569\,\,\times \,\,{{10}^{-10}})}^{3}}}\] \[=\,\,\,\,3.506 gc{{m}^{-}}^{3}\]
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