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: N _ A =6× 10^ 23 , atomtextc weight (in g/mol ) TI=204,Al=27,S=32,O=16,H=1 )

Correct answer: C. 47 For any unit cell d= 2M a^ 3 N _ A (for BCC,Z=2 ) 2.32 = 2 × M 6 × 1 0 ^ 23 × (1.22) ^ 3 × 1 0 ^ – 21 ∴ M = 1264 So 4 2 3 + x (1 8) = 1 2 6 4 & x = 4 7

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Solid State

NTA Abhyas 2020 · Chemistry · Solid State

TlAl ( SO_ 4 ) _ 2 . xH_ 2 O is bcc with edge length ‘a’ = 1.22nm ....

\text{T}\text{l}\text{A}\text{l}{\left({\text{S}\text{O}_{4}\right)}_{2}. {\text{x}\text{H}_{2}\text{O}

is bcc with edge length

\text{‘}\text{a}\text{’} = 1.22\text{n}\text{m}

. If the density of the solid is

2.32\text{g} / \text{c}\text{c}

, then the value of

\text{x}

is
(Given:

{N}_{A} \text{=}\text{6}\text{\times}{\text{10}^{\text{23}} \text{,}

atomtextc weight (in

\text{g/mol}

)

\text{TI}\text{=}\text{204,}\text{Al}\text{=}\text{27,}\text{S}\text{=}\text{32,}\text{O}\text{=}\text{16,}\text{H}\text{=}\text{1}

)

Options

Correct answer

C. 47

Step-by-step solution

For any unit cell

\text{d}\text{=}\dfrac{\text{2}\text{M}}{a^{3} {N}_{A}}

(for

\text{BCC,}\text{Z}\text{=}\text{2}

)

2.32 = \dfrac{2 \times M}{6 \times 1{0}^{23} \times {\left(1.22\right)}^{3} \times 1{0}^{– 21}}

∴ M = 1264

4 2 3 + \text{x} \left(1 8\right) = 1 2 6 4

\text{x} = 4 7

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