Question
An element has a body-centred cubic bcc structure with a cell edge of $288\ pm$. The density of the element is $7.2g/ cm^3$. How many atoms are present in $208\ g$ of the element?
✓
Answer
Density of unit cell, $\text{d}=\frac{\text{z}\times\text{M}}{\text{a}^3\times\text{N}_{\text{A}}}\ \text{or}\ \text{M}=\frac{\text{d}\times\text{a}^3\times\text{N}_{\text{A}}}{\text{z}}...(\text{i)}$
Here, $z = 2, d = 7.2g/ cm^{-3}, N_A = 6.022 \times 1023mol^{-1},$
$a = 288pm = 288 \times 10^{-10}cm = 2.88 \times 10^{-8}cm,$
Substituting these values in expression (i), we get,
$\text{M}=\frac{7.2\text{g/cm}^{-3}\times(2.88\times10^{-8}\text{cm})^3\times6.022\times10^{23}\text{mol}^{-1}}{2}=51.75\text{g/mol}^{-1}$
Moles of element $=\frac{\text{Masso felement}}{\text{Malarmass}}=\frac{208\text{g}}{51.78\text{g/mol}^{-1}}=4.02\text{mol}$
$\therefore$ toms present in 208g of element $= 6.022 \times 10^{23} \times 4.02$ atoms,
$2.421 \times 10^{24}$ atoms
Here, $z = 2, d = 7.2g/ cm^{-3}, N_A = 6.022 \times 1023mol^{-1},$
$a = 288pm = 288 \times 10^{-10}cm = 2.88 \times 10^{-8}cm,$
Substituting these values in expression (i), we get,
$\text{M}=\frac{7.2\text{g/cm}^{-3}\times(2.88\times10^{-8}\text{cm})^3\times6.022\times10^{23}\text{mol}^{-1}}{2}=51.75\text{g/mol}^{-1}$
Moles of element $=\frac{\text{Masso felement}}{\text{Malarmass}}=\frac{208\text{g}}{51.78\text{g/mol}^{-1}}=4.02\text{mol}$
$\therefore$ toms present in 208g of element $= 6.022 \times 10^{23} \times 4.02$ atoms,
$2.421 \times 10^{24}$ atoms
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