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Standard XII

Chemistry

Density

Iron exhibits...

Question

Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is:

\frac{3 (\sqrt{3})}{4 (\sqrt{2})}

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\frac{4 (\sqrt{3})}{3 (\sqrt{2})}

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\frac{(\sqrt{3})}{(\sqrt{2})}

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\frac{1}{2}

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Solution

The correct option is A $\frac{3 (\sqrt{3})}{4 (\sqrt{2})}$
For BCC, z = 2 and a = $\frac{4 r}{(\sqrt{3})}$
For FCC, z = 4 and a = 2 $\sqrt{2}$ r
$∴ \frac{d_{25^{\circ} C}}{d_{900^{\circ} C}} = \frac{{(\frac{Z M}{N_{A} a^{3}})}_{B C C}}{{(\frac{Z M}{N_{A} a^{3}})}_{F C C}}$
$= \frac{2}{4} {(\frac{2 \sqrt{2} r}{\frac{4 r}{\sqrt{3}}})}^{3}$
$= (\frac{3 \sqrt{3}}{4 \sqrt{2}})$

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Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is:

The correct option is A 3 (√3)4 (√2)For BCC, z = 2 and a = 4r (√3) For FCC, z = 4 and a = 2 √2r ∴d25∘Cd900∘C=(ZMNAa3)BCC(ZMNAa3)FCC =24(2√2r4r√3)3 =(3√34√2)