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$TlAl {({SO}_{4})}_{2} \cdot {xH}_{2} O is bcc with a = 1 .22 nm . Ifthe density of the solid is 2 .32 g / cc, then the value of x is (Given : N_{A} = 6 \times 10^{23}; atomic weight: Tl = 204, Al = 27, S = 32, O = 16, H = 1)$

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answer is C.

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Detailed Solution

$2 .32 = \frac{2 \times M}{6 \times 10^{23} \times (1 .22)^{3} \times 10^{- 21}}$

$∴ M = 1264 and x ≅ 47$

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