The unit cell of copper corresponds to a face centered cube of edge length 3.596 $\mathop A\limits^o$ with one copper atom at each lattice point. The calculated density of copper in kg/m3 is ___________. [Molar mass of Cu : 63.54 g; Avogadro Number = 6.022 $\times$ 1023]
Answer (integer)
9077
Solution
<p>Density of copper, $d = {{Z \times M} \over {{a^3} \times {N_A}}}$</p>
<p>Given, Z = 4, for fcc lattice,</p>
<p>M = 63.54 g mol<sup>$-$1</sup></p>
<p>= 63.54 $\times$ 10<sup>$-$3</sup> kg mol<sup>$-$1</sup>,</p>
<p>a = 3.596 $\mathop A\limits^o$ = 3.596 $\times$ 10<sup>$-$10</sup> m,</p>
<p>N<sub>A</sub> = 6.022 $\times$ 10<sup>23</sup> mol<sup>$-$1</sup></p>
<p>On putting given values, we get</p>
<p>$$ \Rightarrow d = {{4 \times (63.54 \times {{10}^{ - 3}})} \over {{{(3.596 \times {{10}^{ - 10}})}^3} \times (6.022 \times {{10}^{23}})}}$$ kg/m<sup>3</sup></p>
<p>$= 9076.26 \simeq 9077$ kg/m<sup>3</sup></p>
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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