The number of atoms in 100 g of an fcc crystal with density d - 10 g/cm3 and cell edges as 200 pm is equal to

M=d×a3×NAn=10×200×10−33×6.02×10234=12.04g Number of atoms in 100g=6.02×102312.04×100=5×1024 .

The number of atoms in 100 g of an fcc crystal with density d - 10 g/cm3 and cell edges as 200 pm is equal to

M=d×a3×NAn=10×200×10−33×6.02×10234=12.04g Number of atoms in 100g=6.02×102312.04×100=5×1024 .

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The number of atoms in 100 g of an fcc crystal with density d - 10 g/cm³ and cell edges as 200 pm is equal to

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$3 \times 10^{25}$

$5 \times 10^{24}$

$1 \times 10^{25}$

$5.96 \times 10^{- 3}$

answer is B.

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$M = \frac{d \times a^{3} \times N_{A}}{n}$

$= \frac{10 \times {(200 \times 10^{- 3})}^{3} \times (6.02 \times 10^{23})}{4} = 12.04 g$

$Number of atoms in 100 g = \frac{6.02 \times 10^{23}}{12.04} \times 100 = 5 \times 10^{24} .$

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The number of atoms in 100 g of an fcc crystal with density d - 10 g/cm3 and cell edges as 200 pm is equal to

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The number of atoms in 100 g of an fcc crystal with density d - 10 g/cm³ and cell edges as 200 pm is equal to