Question

When heated above ${916}^{\circ }\text{C}$, iron changes its BCC crystalline form to FCC without any change in the radius of atom. The ratio of density of the crystal before heating and after heating is [atomic weight of $Fe=56$]

• A. 1.069
• B. 0.918
• C. 0.725
• D. 1.231

Answer 1

The correct option is B 0.918

We know that,
Density = d =$\frac{Z\times MM}{a^3\times NA}$
The ratio of density of the crystal before heating and after heating is
$\frac{d_1}{d_2}=\frac{\frac{Z_{BCC}\times MM}{a_{BCC}^3\times NA}}{\frac{Z_{FCC}\times MM}{a_{FCC}^3\times NA}}$
$=\frac{Z_{BCC}}{a_{BCC}^3}\frac{a_{FCC}^3}{Z_{FCC}}$
$=\frac{2}{(\frac{4r}{\sqrt{3})}{)}^3}\times \frac{(2\sqrt{2}r{)}^3}{4}$
= 0.918