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JEE Advanced 2017Paper 1
Exam
JEE Advanced
Year
2017
Paper
Paper 1
Subject
Chemistry
Chapter
Solid State

JEE Advanced 2017 · Paper 1 · Chemistry · Solid State

A crystalline solid of a pure substance has a face-centred cubic...

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400pm400pm . If the density of the substance in the crystal is 8gcm−3,8g{cm}^{- 3}, then the number of atoms present in 256g256g of the crystal isN×1024isN \times {10}^{24} . The value of NNis

Correct answer

0

Step-by-step solution

Formula of density =Z×MNA×a3= \dfrac{Z \times M}{{N}_{A} \times {a}^{3}}

For FCC unit cell Z,=4Z, = 4

Edge length, a=4×10−8cma = 4 \times {10}^{- 8}cm

M=d×NA×a3Z=8×6×1023×64×10−244g/molM = \dfrac{d \times {N}_{A} \times {a}^{3}}{Z} = \dfrac{8 \times 6 \times {10}^{23} \times 64 \times {10}^{- 24}}{4}g / mol

Number of atoms =wt(g)molarmass×NA=256×10×6×10238×6×16=2×1024= \dfrac{wt\left(g\right)}{molarmass} \times {N}_{A} = \dfrac{256 \times 10 \times 6 \times {10}^{23}}{8 \times 6 \times 16} = 2 \times {10}^{24}

Hence, the required value of N=2N = 2

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