An element with density 6 {\g} {\cm}^{-3} forms a fcc lattice with edge length of 4 \times 10^{-8} {\cm}. The molar mass of the element is ( {\N}_{ {\A}}=6 \times 10^{23} {\mol}^{-1})

Correct Answer is : 57.6gmol−1

CBSE Class 12 Chemistry 2022 Term 1 Paper - Question 32

Question : 32 of 55

Marks: +1, -0

An element with density

6 {\g} {\cm}^{-3}

forms a fcc lattice with edge length of

4 × 10^{-8} {\cm}

. The molar mass of the element is

( {\N}_{ {\A}}=6 × 10^{23} {\mol}^{-1})

Solution:

Number of atoms per unit cell of FCC

=4

Z=4

We have given -

{\d}=6 {\g} \/ {\cm}^3,

{\a}=\;\text " edge length "\;=4 × 10^{-8} {\cm}

{\N}_{ {\A}}=6 × 10^{23}

{\d}=\;{Z M}/{a^3 N_A}

{\M}=\;\text " molar mass of element. "\;

∴ 6 {\g} \/ {\cm}^3=\;{4 × M}/{ (4 × 10^{-8}) ^3 {\cm}^3 × 6 × 10^{23} {\mol}^{-1}}

⇒ 6 × (64 × 10^{-24}) × 6 × 10^{23}=4 {\M}

⇒ {\M}=\;{6 × 6 × 64 × 10^{-1}}/{4} {\g} \/ {\mol}^2

∴ {\M}=57.6 {\g} \/ {\mol}^2

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