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TlAl(SO₄)₂. xH₂O is bcc with 'a' = 1.22 nm. If the density of the solid is 2.32 g/cc, then the value of x is (Given: N_{A =} 6 x 10²³; at. mass: Tl = 204, Al = 27, S = 32).

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answer is C.

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Detailed Solution

$2.32 = \frac{2 \times M w}{6 \times 10^{23} {(1.22)}^{3} \times 10^{- 21}} (1 n m = 10^{- 9} m = 10^{- 7} c m) \Rightarrow M w = 1264 T I A I {(S O_{4})}_{2} . x H_{2} O = 1264 204 + 27 + 2 \times 96 + x \times 18 = 1264 x = 46.72 \approx 47$

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