An element with density 10 gcm -3 forms a cubic unit cell with edge length of 3 - 10-8 cm- What is the nature of the cubic unit cell if the atomic mass of the element is 81 g mol -1 -

Z =ρ× a^3 × N_0/M =10g cm^ 3× (3 × 10^ 8 cm)^3 (6.023 × 10^23 mol^ 1)/81 g mol^ 1=2 Hence, the element has body centred cubic unit cell. ...

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Standard XII

Chemistry

Crystalline and Amorphous Solid

An element wi...

Question

An element with density $10 g c m^{- 3}$ forms a cubic unit cell with edge length of $3 \times 10^{- 8}$ cm. What is the nature of the cubic unit cell if the atomic mass of the element is $81 g m o l^{- 1}$ ?

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Solution

$Z = \frac{ρ \times a^{3} \times N_{0}}{M}$

$= \frac{10 g c m^{- 3} \times (3 \times 10^{- 8} c m)^{3} (6.023 \times 10^{23} m o l^{- 1})}{81 g m o l^{- 1}} = 2$

Hence, the element has body centred cubic unit cell.

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An element with density 10 gcm−3 forms a cubic unit cell with edge length of 3×10−8 cm. What is the nature of the cubic unit cell if the atomic mass of the element is 81 g mol−1?

Z=ρ×a3×N0M =10gcm−3×(3×10−8cm)3(6.023×1023mol−1)81 g mol−1=2 Hence, the element has body centred cubic unit cell.

An element with density $10 g c m^{- 3}$ forms a cubic unit cell with edge length of $3 \times 10^{- 8}$ cm. What is the nature of the cubic unit cell if the atomic mass of the element is $81 g m o l^{- 1}$ ?

$Z = \frac{ρ \times a^{3} \times N_{0}}{M}$

$= \frac{10 g c m^{- 3} \times (3 \times 10^{- 8} c m)^{3} (6.023 \times 10^{23} m o l^{- 1})}{81 g m o l^{- 1}} = 2$

Hence, the element has body centred cubic unit cell.