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Q.

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: NA = 6 x 1023; at. mass: Tl = 204, Al = 27, S = 32).

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a

2

b

4

c

47

d

70

answer is C.

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Detailed Solution

2.32=2×Mw6×10231.223×10-21 ( 1nm=10-9m=10-7cm) ⇒Mw=1264 TIAI(SO4)2.xH2O = 1264 204+27+2×96+x×18 =1264 x=46.72≈47

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