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{C}{s}{B}{r} crystallises in a body- centred cubic lattice. Th

Question

$C s B r$ crystallises in a body $-$ centred 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's number being $6.02 \times 10^{23} m o l^{- 1}$ , the density of $C s B r$ is $:$

42.5 g / c m^{3}

0.425 g / c m^{3}

8.25 g / c m^{3}

4.25 g / c m^{3}

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Amelia Discussed

C s B r

crystallises in a body

-

centred 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's number being

6.02 \times 10^{23} m o l^{- 1}

, the density of

C s B r

:

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Discuss this question LIVE

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Text solutionVerified

Density of CsBr = \frac{Z \times M}{a ^{3} \times N _{0}}

Z \to number of atoms in the bcc unit cell = 2

M \to molar mass of CsBr = 133 + 80 = 213

a \to edge length of unit cell = 436.6 pm

= 436.6 \times 10^{- 10} c m

∴ Density = \frac{2 \times 213}{( 436.6 \times 10 ^{- 10} ) ^{3} \times 6.023 \times 10 ^{23}}

= 8.49 \times 10^{- 7} \times 10^{7} g / c m^{3}

= 8.50 g / c m^{3}

For a unit cell

= \frac{8.50}{2} = 4.25 g / c m^{3}

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Question Text	$C s B r$ crystallises in a body $-$ centred 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's number being $6.02 \times 10^{23} m o l^{- 1}$ , the density of $C s B r$ is $:$
Answer Type	Text solution:1
Upvotes	150