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$The number of atoms in 100 g of an fcc crystal with density, d = 10 g / {cm}^{3} and cell edge equal to 100 pm, is equal to$

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$1 \times 10^{25}$

$2 \times 10^{25}$

$3 \times 10^{25}$

$4 \times 10^{25}$

answer is D.

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Detailed Solution

$M = \frac{ρ \times a^{3} \times N_{A} \times 10^{- 30}}{Z} = \frac{10 \times (100)^{3} \times 6 .02 \times 10^{23} \times 10^{- 30}}{4} = 15 .05 ∴ Number of atoms in 100 g = \frac{6 .02 \times 10^{23}}{15 .05} \times 100 = 4 \times 10^{25}$

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