SATHEE: The Solid State - Result Question 12

The Solid State - Result Question 12

####12. Iron exhibits $b c c$ structure at room temperature. Above $900^{\circ} C$ , it transforms to $f c c$ 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

[2018]

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

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

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

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Solution:

(d) For bcc lattice: $Z = 2, a = \frac{4 r}{\sqrt{3}}$

For fcc lattice : $Z = 4, a = 2 \sqrt{2} r$

$\begin{matrix} ∴ \frac{d_{25^{\circ} C}}{d_{900^{\circ} C}} = \frac{(\frac{Z M}{N_{A} a^{3}})_{b c c}}{(\frac{Z M}{N_{A} a^{3}})_{f c c}} \\ = \frac{2}{4} (\frac{2 \sqrt{2} r}{\frac{4 r}{\sqrt{3}}})^{3} = \frac{3 \sqrt{3}}{4 \sqrt{2}} \end{matrix}$

The Solid State - Result Question 12

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The Solid State - Result Question 12

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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