SATHEE: The Solid State - Result Question 25

The Solid State - Result Question 25

####25. $C s B r$ crystallises in a body centered cubic lattice. The unit cell length is $436.6 p m$ . Given that the atomic mass of $C s = 133$ and that of $B r = 80 a m u$ and Avogadro number being $6.02 \times 10^{23} m o l^{- 1}$ , the density of $C s B r$ is

(a) $0.425 g / c m^{3}$

(b) $8.25 g / c m^{3}$

(d) $42.5 g / c m^{3}$

[2006]

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Solution:

Atomic mass of unit cell $= 133 + 80 = 213$ a.m.u

Volume of cell $= (436.6 \times 10^{- 10})^{3} c m^{3}$

$\begin{aligned} Density & = \frac{Z \times at.wt.}{Av.no. \times vol.of unit cell} \\ Density & = \frac{1 \times 213}{6.02 \times 10^{23} \times (436.6)^{3} \times 10^{- 30}} \\ = \frac{213 \times 10^{7}}{6.02 \times (436.6)^{3}} = 4.25 g / c m^{3} \end{aligned}$

In the density formula, $Z$ is number of particles present in a unit cell and $M$ is molar mass of that particle. For any diatomic molecule like $C s B r, Z =$ 2 but for any mono atomic molecule like $L i, Z = 1$ for bec lattice.

The Solid State - Result Question 25

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The Solid State - Result Question 25

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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