Question

The molar volume of KCl and NaCl are 37.46 ml and 27.94 ml. The ratio of unit cell edges of the two crystals is :

• A. 1.296
• B. 1.116
• C. 1.341
• D. 0.950

Answer 1

The correct option is B 1.116

Molar volume $=N_A\times \frac{4}{3}\pi r^3$ where $N_A$ is Avogadro number,
Now Molar volume of $KCl$ is $37.46c{m}^3$ So $N_A\times \frac{4}{3}\pi R^3=37.46c{m}^3$
$6.022\times {10}^{23}\times \frac{4}{3}\pi R^3=37.46c{m}^3$
On solving this gives $R=2.458\times {10}^{-8}cm$
And for $NaCl$ by using the same formula it gives $R=2.22\times {10}^{-8}cm$
Now the ratio of edge length must be equal to ratio of radii as both belong to same crystal system so $\frac{KCl}{NaCl}=\frac{2.458}{2.22}=1.108$