JEE Main Chemistry question

Not in JEE Main or NEET syllabus

A fcc element (atomic mass = 60) has a cell edge of 400 pm. Its density is:

Answer: (A)

Solution

For an fcc lattice, , so \dfrac{ZM}{N_Aa^3}\dfrac{4×60}{6.02×10^{23}×(4×10^{-8})^3}\dfrac{240}{6.02×10^{23}×6.4×10^{-23}}\dfrac{240}{38.53}$

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Question ID VK-141783