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Iron exhibits bcc structure at room temperature. Above $900^\circ C$, it transforms to fcc structure. The ratio of density of iron at room temperature to that at $900^\circ C$ (assuming molar mass and atomic radii of iron remains constant with temperature) is
Explanation
Use $d = ZM/N_Aa^3$ with the edge–radius relation for each lattice.
Detailed Solution
For BCC lattice: Z = 2, $a = \frac{4r}{\sqrt{3}}$
For FCC lattice: Z = 4, $a = 2\sqrt{2}\,r$
$\therefore \frac{d_{25^\circ C}}{d_{900^\circ C}} = \frac{\left(\frac{ZM}{N_Aa^3}\right)_{BCC}}{\left(\frac{ZM}{N_Aa^3}\right)_{FCC}}$
$= \frac{2}{4}\left(\frac{2\sqrt{2}\,r}{\frac{4r}{\sqrt{3}}}\right)^3$
$= \frac{3\sqrt{3}}{4\sqrt{2}}$
For FCC lattice: Z = 4, $a = 2\sqrt{2}\,r$
$\therefore \frac{d_{25^\circ C}}{d_{900^\circ C}} = \frac{\left(\frac{ZM}{N_Aa^3}\right)_{BCC}}{\left(\frac{ZM}{N_Aa^3}\right)_{FCC}}$
$= \frac{2}{4}\left(\frac{2\sqrt{2}\,r}{\frac{4r}{\sqrt{3}}}\right)^3$
$= \frac{3\sqrt{3}}{4\sqrt{2}}$