Copper crystallises in fcc unit cell with cell edge length of 3-608 - 10-8 cm- The density of copper is 8-92 g cm-3- Calculate the atomic mass of copper-

D=Z × Ma^3 × N_A 8.92 g cm^ 3=4 × M(3.608 × 10^ 8)^3 × 6.023 × 10^23 M=8.92 g cm^ 3× (3.608 × 10^ 8)^3 cm^3 × 6.023 × 10^234 M=8.92 × 46.96 × 6.023 × 10^ 1 g4 ...

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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

Copper crysta...

Question

Copper crystallises in fcc unit cell with cell edge length of $3.608 \times 10^{- 8} cm$ . The density of copper is $8.92 {g cm}^{- 3}$ . Calculate the atomic mass of copper.

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Solution

$d = \frac{Z \times M}{a^{3} \times N_{A}}$
$8.92 {g cm}^{- 3} = \frac{4 \times M}{(3.608 \times 10^{- 8})^{3} \times 6.023 \times 10^{23}}$
$M = \frac{8.92 {g cm}^{- 3} \times (3.608 \times 10^{- 8})^{3} {cm}^{3} \times 6.023 \times 10^{23}}{4}$
$M = \frac{8.92 \times 46.96 \times 6.023 \times 10^{- 1} g}{4}$
$M = \frac{8.92 \times 4.686 \times 6.023}{4} = 63.07 g/mol$
Atomic mass of $C u = 63.07 u$
$\approx 63.1 u$

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Copper crystallises in fcc unit cell with cell edge length of 3.608×10–8 cm. The density of copper is 8.92 g cm–3. Calculate the atomic mass of copper.

d=Z×Ma3×NA 8.92 g cm−3=4×M(3.608×10−8)3×6.023×1023 M=8.92 g cm−3×(3.608×10−8)3 cm3×6.023×10234 M=8.92×46.96×6.023×10−1 g4 M=8.92×4.686×6.0234=63.07 g/mol Atomic mass of Cu=63.07 u ≈63.1 u

Copper crystallises in fcc unit cell with cell edge length of $3.608 \times 10^{- 8} cm$ . The density of copper is $8.92 {g cm}^{- 3}$ . Calculate the atomic mass of copper.