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Q.

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is $8 g c m^{- 3}$ , then the number of atoms present in 256 g of the crystal is $N \times 10^{24}$ . The value of N is

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Detailed Solution

$\begin{array}{l} FCC: Z = 4 \\ d = \frac{Z \times M}{a^{3} \cdot N_{\circ}} \\ 8 = \frac{4 \times (Mol. Wt)}{(400 \times 10^{- 10})^{3} \times N_{0}} \\ Mol . Wt = 77 g / mol \end{array}$

$\begin{array}{l} no . of atoms in 265 g = \frac{256 g}{77 g / mol} \times 6.022 \times 10^{23} = \underset{N}{\underset{⏟}{2}} \times 10^{24} \end{array}$

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