JEE Advance - Chemistry (2017 - Paper 1 Offline - No. 7)
A crystalline solid of a pure substance has a face-centered 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
Answer
2
Explanation
Given:
(i) Edge length of face centred cubic (FCC) unit cell (a) $=400 \mathrm{pm}$
(ii) Density of the substance $(\rho)=8 \mathrm{~g} \mathrm{~cm}^{-3}$
(iii) Mass of the crystal $(m)=256 \mathrm{~g}$
To Find: The value of $\mathrm{N}$ in $\mathrm{N} \times 10^{24}$
Formula: Density of the cell $(\rho)$
$$ \begin{aligned} & =\frac{\text { Mass of crystal }(\mathrm{M}) \times \begin{array}{l} \text { No. of atoms in unit cell } \\ \end{array}}{\text { Volume of unit cell } \times \text { No. of atoms of pure substance in mass m }} \\ \end{aligned} $$
Calculations: A face-centred cubic (F.C.C.) lattice has atoms of pure substance at the corner of the unit cell and/or atom each at the centre of the face of the unit cell.
Since there are eight atoms at the corners and one atom at each of the six faces (i.e., six atoms in total):
Total number of atoms present per unit cell $(z)$
$$ =\frac{1}{8} \times 8+\frac{1}{2} \times 6=4 $$
Density of the cell $(\rho)$
$$ \begin{aligned} & =\frac{256 \mathrm{~g} \times 4}{\left(400 \times 10^{-10}\right)^3 \times \mathrm{N}^1} \\\\ & 8 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{256 \mathrm{~g} \times 4}{\left(400 \times 10^{-10}\right) \times \mathrm{N}^1} \\\\ & \mathrm{~N}^{\prime}=\frac{256 \mathrm{~g} \times 4}{8 \mathrm{~g} \mathrm{~cm}^{-3} \times(400)^3 \times 10^{-30}} \\\\ & \mathbf{N}^{\prime}=2 \times 10^{24} \text { atoms } \end{aligned} $$
Hence, the crystal of pure substance of mass $256 \mathrm{~g}$ contains $2 \times 10^{24}$ atoms.
Now, $\mathrm{N}^{\prime}=\mathrm{N} \times 10^{24}$
Therefore, value of $\mathrm{N}=2$.
(i) Edge length of face centred cubic (FCC) unit cell (a) $=400 \mathrm{pm}$
(ii) Density of the substance $(\rho)=8 \mathrm{~g} \mathrm{~cm}^{-3}$
(iii) Mass of the crystal $(m)=256 \mathrm{~g}$
To Find: The value of $\mathrm{N}$ in $\mathrm{N} \times 10^{24}$
Formula: Density of the cell $(\rho)$
$$ \begin{aligned} & =\frac{\text { Mass of crystal }(\mathrm{M}) \times \begin{array}{l} \text { No. of atoms in unit cell } \\ \end{array}}{\text { Volume of unit cell } \times \text { No. of atoms of pure substance in mass m }} \\ \end{aligned} $$
Calculations: A face-centred cubic (F.C.C.) lattice has atoms of pure substance at the corner of the unit cell and/or atom each at the centre of the face of the unit cell.
Since there are eight atoms at the corners and one atom at each of the six faces (i.e., six atoms in total):
Total number of atoms present per unit cell $(z)$
$$ =\frac{1}{8} \times 8+\frac{1}{2} \times 6=4 $$
Density of the cell $(\rho)$
$$ \begin{aligned} & =\frac{256 \mathrm{~g} \times 4}{\left(400 \times 10^{-10}\right)^3 \times \mathrm{N}^1} \\\\ & 8 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{256 \mathrm{~g} \times 4}{\left(400 \times 10^{-10}\right) \times \mathrm{N}^1} \\\\ & \mathrm{~N}^{\prime}=\frac{256 \mathrm{~g} \times 4}{8 \mathrm{~g} \mathrm{~cm}^{-3} \times(400)^3 \times 10^{-30}} \\\\ & \mathbf{N}^{\prime}=2 \times 10^{24} \text { atoms } \end{aligned} $$
Hence, the crystal of pure substance of mass $256 \mathrm{~g}$ contains $2 \times 10^{24}$ atoms.
Now, $\mathrm{N}^{\prime}=\mathrm{N} \times 10^{24}$
Therefore, value of $\mathrm{N}=2$.
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