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NCERT Class XII Chemistry
Chapter - The Solid State
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Question : 13 of 50
Marks: +1, -0
13.Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm−3^{-3}, calculate atomic radius of niobium using its atomic mass 93 u.
Solution:  
Given d=8.55  g  cm−3,M=93  g  mol−1,Z=2d=8.55\;\mathrm{g}\;\mathrm{cm}^{-3}, M=93\;\mathrm{g}\;\mathrm{mol}^{-1}, Z=2 (for bcc) ,NA=6.022×1023,r=?, N_{A}=6.022 \times 10^{23}, r = ?
Using formula
a3=M×Zd×NAa^{3}= \frac{M \times Z}{d \times N_{A}}
=93×28.55×6.022×1023= \frac{93 \times 2}{8.55 \times 6.022 \times 10^{23}}
⇒a3=3.61×10−23\Rightarrow a^{3}=3.61 \times 10^{-23}
              =36.1×10−24\;\;\;\;\;\;\;=36.1 \times 10^{-24}
∴      a=3.304×10−8  cm\therefore\;\;\; a=3.304 \times 10^{-8} \;\mathrm{cm}
                =330.4×10−12  m\;\;\;\;\;\;\;\;=330.4 \times 10^{-12} \;\mathrm{m}
                =330.4  pm\;\;\;\;\;\;\;\;=330.4 \;\mathrm{pm}
For body-centred cubic,
r=34ar=\frac{\sqrt{3}}{4} a
=0.433a=0.433 a
=0.433×330.4  pm=0.433 \times 330.4 \;\mathrm{pm}
=143.1  pm=143.1 \;\mathrm{pm}
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