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CBSE Class 12 Chemistry 2022 Term 1 Paper

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Question : 32 of 55
Marks: +1, -0
An element with density 6 g cm−36\ \mathrm{g}\ \mathrm{cm}^{-3} forms a fcc lattice with edge length of 4×10−8 cm4 \times 10^{-8}\ \mathrm{cm}. The molar mass of the element is (NA=6×1023 mol−1)\left( N_A = 6 \times 10^{23}\ \mathrm{mol}^{-1} \right)
Solution:  
Number of atoms per unit cell of FCC =4=4
Z=4Z=4
We have given -
d=6 g/cm3,d = 6\ \mathrm{g} / \mathrm{cm}^3,
a= edge length =4×10−8 cma = \text{ edge length } = 4 \times 10^{-8}\ \mathrm{cm}
NA=6×1023N_A = 6 \times 10^{23}
d=ZMa3NAd = \frac{Z M}{a^3 N_A}
M= molar mass of element. M = \text{ molar mass of element. }
∴6 g/cm3=4×M(4×10−8)3 cm3×6×1023 mol−1\therefore 6\ \mathrm{g} / \mathrm{cm}^3 = \frac{4 \times M}{(4 \times 10^{-8})^3\ \mathrm{cm}^3 \times 6 \times 10^{23}\ \mathrm{mol}^{-1}}
⇒6×(64×10−24)×6×1023=4M\Rightarrow 6 \times (64 \times 10^{-24}) \times 6 \times 10^{23} = 4M
⇒M=6×6×64×10−14 g/mol2\Rightarrow M = \frac{6 \times 6 \times 64 \times 10^{-1}}{4}\ \mathrm{g} / \mathrm{mol}^2
∴M=57.6 g/mol2\therefore M = 57.6\ \mathrm{g} / \mathrm{mol}^2
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