TlAl((SO)4)2 . (xH)2O is bcc with edge length a= 1.22 nm . If the density of the solid is 2.32 g/cc , then the value of x is Given: NA=6× 1023 Atomic weight (in g/mol ) TI=204, Al=27, S=32, O=16, H=1 )

Question

TlAl((SO)4)2 . (xH)2O is bcc with edge length a= 1.22 nm . If the density of the solid is 2.32 g/cc , then the value of x is Given: NA=6× 1023 Atomic weight (in g/mol ) TI=204, Al=27, S=32, O=16, H=1 ) (A) 2 (B) 4 (C) 47 (D) 70

Tardigrade Admin · Accepted Answer

For any unit cell textd text= ( text2 textM/ texta text3 textN textA) (for textBCC, textZ text= text2 ) 2.32=(2 × M/6 × 1 023 × (1.22)3 × 1 0- 21) ∴ M=1264 So 4 2 3 + textx (1 8) = 1 2 6 4 textx = 4 7

Q. $Tl A l ((SO)_{4})_{2} . (x H)_{2} O$ is bcc with edge length $a = 1.22 nm$ . If the density of the solid is $2.32 g / cc$ , then the value of $x$ is
Given: $N_{A} = 6 \times 1 0^{23}$
Atomic weight (in $g / m o l$ ) $T I = 204, A l = 27, S = 32, O = 16, H = 1$ )

$2$

$4$

$47$

$70$

For any unit cell
$d = \frac{2 M}{a ^{3} N _{A}}$ (for $BCC, Z = 2$ )
$2.32 = \frac{2 \times M}{6 \times 1 0 ^{23} \times ( 1.22 ) ^{3} \times 1 0 ^{- 21}}$
$∴ M = 1264$
So $423 + x (18) = 1264$
$x = 47$

Q. TlAl((SO)4​)2​.(xH)2​O is bcc with edge length a=1.22nm . If the density of the solid is 2.32g/cc , then the value of x is Given: NA​=6×1023 Atomic weight (in g/mol ) TI=204,Al=27,S=32,O=16,H=1 )

2

4

47

70

For any unit cell d=a3NA​2M​ (for BCC,Z=2 ) 2.32=6×1023×(1.22)3×10−212×M​ ∴M=1264 So 423+x(18)=1264 x=47

Q. $Tl A l ((SO)_{4})_{2} . (x H)_{2} O$ is bcc with edge length $a = 1.22 nm$ . If the density of the solid is $2.32 g / cc$ , then the value of $x$ is
Given: $N_{A} = 6 \times 1 0^{23}$
Atomic weight (in $g / m o l$ ) $T I = 204, A l = 27, S = 32, O = 16, H = 1$ )

$2$

$4$

$47$

$70$

For any unit cell
$d = \frac{2 M}{a ^{3} N _{A}}$ (for $BCC, Z = 2$ )
$2.32 = \frac{2 \times M}{6 \times 1 0 ^{23} \times ( 1.22 ) ^{3} \times 1 0 ^{- 21}}$
$∴ M = 1264$
So $423 + x (18) = 1264$
$x = 47$