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A metal having atomic mass $60.23 gm / mole$ crystallises, in ABCABC close packing. Calculate the density of each metal atom if edge length is $10 \overset{°}{A}$
[Given : $N_{A} = 6 .023 \times 10^{23}$ ]

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0.4 gm/ml

40 gm/ml

0.54 gm/ml

54 gm/ml

answer is C.

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Detailed Solution

$\begin{aligned} a = 10 A^{\circ} = 10 \times 10^{- 8} cm \\ ABCABC \Rightarrow FCC \Rightarrow Z = 4 \end{aligned}$

$\begin{array}{l} d = \frac{2 \times M}{a^{3} \times N_{A}} \\ N_{A} = 6 .023 \times 10^{23} \end{array}$

$ρ = \frac{4 \times 60 .23}{{(10 \times 10^{- 8})}^{3} \times N_{A}}$

$\begin{array}{l} d = \frac{4 \times 60 .23}{{(10 \times 10^{- 8})}^{3} \times 6 .023 \times 10^{23}} \\ d = \frac{4 \times 60 .23}{10^{- 21} \times 10^{23} \times 6 .023} \end{array}$

$\begin{array}{l} d = \frac{4 \times 10}{100} = 0 .4 \\ d = 0 .4 g / {cm}^{3} \end{array}$

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