First slide

Crystal lattices and unit cells

Question

An element crystallises as face centred cubic lattice with density as 5.20 g / cm³ and edge length of the side of unit cell as 300 pm. The mass of the element which contains 3.0×10²⁴ cannot be equal to

Moderate

A

55g
B

105g
C

205g
D

275g

Solution

Z = 4 in fcc lattice M = ?. d = 5.20 g / cm³
a = 300, pm = 300 × 10^-12. m = 3 × 10^-10. m = 3 × 10^-6 cm³
$∴ a^{3} = 27 \times 10^{- 24} {cm}^{3}$
$N_{0} = 6 .02 \times 10^{23}$
$∴ M = \frac{{dN}_{0} a^{3}}{Z}$
$= \frac{5 .20 g / {cm}^{3} \times 6 .02 \times 10^{23} {mol}^{- 1} \times 27 \times 10^{- 24} {cm}^{3}}{4}$
= 21.13 g mol^-1
Thus, 6.02×10²³ atmos have = 21.13 g
$∴ 3.01 \times 10^{24} at o ms have = \frac{21.13}{6.02 \times 10^{23}} \times 3.01 \times 10^{24} g$
=105.65 g

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