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JEE Main 20199 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Solid State

JEE Main 2019 · 9 January · Evening Shift · Chemistry · Solid State

At 100 ^ o C, copper (Cu) has FCC unit cell structure with cell...

At 100oC,{100}^{o}C, copper \left(Cu\right)hashasFCCunitcellstructurewithcelledgelengthofunit cell structure with cell edge length ofxÅ.Whatistheapproximatedensityof. What is the approximate density ofCu(in(ingc{m}^{- 3})atthistemperature?<br/><br/>[AtomicMassofCu=63.55u]) at this temperature?<br /><br />\left[AtomicMassofCu = 63.55u\right]

Options

  1. A105x3\dfrac{105}{{x}^{3}}
  2. B205x3\dfrac{205}{{x}^{3}}
  3. C422x3\dfrac{422}{{x}^{3}}
  4. D211x3\dfrac{211}{{x}^{3}}

Correct answer

C. 422x3\dfrac{422}{{x}^{3}}

Step-by-step solution

Crystal structure is FCC,FCC, so number of atoms per unit cell \left(N\right) = 4<br/><br/>Edgelength<br /><br />Edge length= x . {Å}^{} = x{10}^{- 8}c{m}^{3}<br/><br/>Density<br /><br />Density\dfrac{ZM}{{N}_{A}{a}^{3}} = \dfrac{4 \times 63 .5}{\left(6 .03 \times {10}^{23}\right){\left(x \times {10}^{- 8}\right)}^{3}}gm / {cm}^{3}<br/><br/><br /><br />= \dfrac{4 \times 63.5}{\left(6.0\right) \times {x}^{3} \times {10}^{- 1}} = \dfrac{422}{{x}^{3}}gm / c{m}^{3} = \dfrac{422}{{x}^{3}}
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