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Q.

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

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a

55g

b

105g

c

205g

d

275g

answer is A.

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Detailed Solution

For fcc lattice:

$Z = 4 N_{\circ} = 6.02 \times 10^{23} a = 300 pm = 3 \times 10^{- 8} cm$

$M = \frac{d \times N_{\circ} \times a^{3}}{Z}$

$M = \frac{5 .20 g / {cm}^{3} \times 6 .02 \times 10^{23} \times {(3 \times 10^{- 8})}^{3} {cm}^{3}}{4} = 21.13 g / mol$

Thus, 6.02×10²³ atoms have = 21.13 g

$3.01 \times 10^{24} atoms have = \frac{21.13}{6.02 \times 10^{23}} \times 3.01 \times 10^{24} g = 105.65 g$

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