# NEET Chemistry The Solid State Questions Solved

BOARD

An element ‘X’ (At. mass = 40 g mol–1) having f.c.c. structure, has unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’.$\left({N}_{A}=6.022×{10}^{23}mo{l}^{-1}\right)$

3mark

Concept Videos :-

#9 | FCC Unit Cell: Packing Fraction
#20 | Rank of HP Unit Cell
#21 | Packing Fraction of HP Unit Cell
#24 | Density of Unit Cell

Concept Questions :-

Density/Formula/packing fraction/ Semiconductors

$d=\frac{zM}{{a}^{3}{N}_{A}}$                                                                                                           1/2mark

$=\frac{4×40}{{\left(4×{10}^{-8}\right)}^{3}×6.022×{10}^{23}}$                                                                                 1/2mark

$=4.15g/c{m}^{3}$                                                                                                      1/2mark

No of unit cells=total no of atoms /4                                                                       1/2mark

=$\left[\frac{4}{40}×6.022×{10}^{23}\right]/4$                                                                   1/2mark

$=1.5×{10}^{22}$                                                                                     1/2mark

(Or any other correct method)

Difficulty Level:

• 58%
• 36%
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