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The number of atoms in \( 2.4 \mathrm{~g} \) of body centred cubic crystal with edge length \( 200 \mathrm{pm} \) is (density
\( =10 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{NA}=6 \times 10^{23} \) atoms \( / \mathrm{mol} \) )
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\( =10 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{NA}=6 \times 10^{23} \) atoms \( / \mathrm{mol} \) )
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Verified Answer
The correct answer is:
\( 6 \times 10^{22} \)
$d=\frac{Z M}{a^{3} N_{A}}$
$M=\frac{d \times a^{3} \times N_{A}}{Z}=\frac{10 \times(200)^{3} \times 10^{-30} \times 6 \times 10^{23}}{2}=24$
$24 \mathrm{~g}$ contains $6 \times 10^{23}$
$2.4 \mathrm{~g}$ contains?
$6 \times 10^{22}$
$M=\frac{d \times a^{3} \times N_{A}}{Z}=\frac{10 \times(200)^{3} \times 10^{-30} \times 6 \times 10^{23}}{2}=24$
$24 \mathrm{~g}$ contains $6 \times 10^{23}$
$2.4 \mathrm{~g}$ contains?
$6 \times 10^{22}$
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