CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm . Given that the atomic mass of Cs = 133 amu and that of Br = 80 a - Chemistry MCQ Chemistry | ExamTest.live

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CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm . Given that the atomic mass of Cs = 133 amu and that of Br = 80 amu and Avogadro number being 6.02 \times 1 0^{23} mol^{- 1}, the density of CsBr is :

A $42.5 g/cm^{3}$
B $0.425 g/cm^{3}$
C $8.25 g/cm^{3}$
D $4.25 g/cm^{3}$

Solution & Step-by-step Explanation

For a bcc lattice of CsBr, there is 1 formula unit per unit cell (Z = 1).Molar mass M = 133 + 80 = 213 g/mol .Density ρ = \frac{Z \times M}{N _{A} \times a ^{3}} = \frac{1 \times 213}{6.02 \times 1 0 ^{23} \times ( 436.6 \times 1 0 ^{- 10} ) ^{3}} . ρ = \frac{213}{6.02 \times 1 0 ^{23} \times 83.2 \times 1 0 ^{- 24}} = \frac{213}{50.08} \approx 4.25 g/cm^{3} .

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CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm . Given that the atomic mass of Cs = 133 amu and that of Br = 80 amu and Avogadro number being 6.02 \times 1 0^{23} mol^{- 1}, the density of CsBr is :

42.5 g/cm^{3}

0.425 g/cm^{3}

8.25 g/cm^{3}

4.25 g/cm^{3}

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CsBr crystallises in a body centred cubic lattice. The unit cell length is $436.6 pm$ . Given that the atomic mass of Cs = $133 amu$ and that of Br = $80 amu$ and Avogadro number being $6.02 \times 1 0^{23} mol^{- 1}$ , the density of CsBr is :

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CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6pm. Given that the atomic mass of Cs = 133amu and that of Br = 80amu and Avogadro number being 6.02×1023mol−1, the density of CsBr is :

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CsBr crystallises in a body centred cubic lattice. The unit cell length is $436.6 pm$ . Given that the atomic mass of Cs = $133 amu$ and that of Br = $80 amu$ and Avogadro number being $6.02 \times 1 0^{23} mol^{- 1}$ , the density of CsBr is :