Shweta thakur Asked a Question
July 7, 2020 7:18 pmpts 30 pts
=6.04 x 103 Ex. 10. The density f a particular crystal of LiF is 2.65 g/cc. X-ray analysis show length of gold atoms that Li' and F 1onS are arranged in a cubic array at a spacing of 2.01 A. From these data calculate the apparent Avogadro costant. Solution: See Example 29, Chapter 1.
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  • Priyanshu kumar Best Answer
    https://youtu.be/DfIKC357Cm4 correlate this with this vieeo explanation
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  • Priyanshu kumar thankyou
    see this
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    Priyanshu kumar
    got shweta here lif is placed on the edge length at a gap of 2.01 A ...
  • Priyanshu kumar thankyou
    see as there is spacing between atoms..so the edge length of cube is not same as the LiF
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  • Ashutosh singh
    No, Shweta.
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    Ashutosh singh
    Some questions just want to understand your depth of knowledge about it..rather than being formula based..we use formula for atoms, not molecules... please note it down! That densi...
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  • Ashutosh singh
    Hey Shweta, refer this!
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    Shweta thakur
    ok sir but sir here can't we use density direct formula