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Problems · Problem 1.6

Q.A compound forms hcp structure. What is the number of

(a) octahedral voids
(b) tetrahedral voids
(c) total voids formed in 0.4 mol of it.
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0.4 mol supplies 2.409×10232.409\times10^{23} atoms; octahedral voids equal that number, tetrahedral voids are twice it, and the total is three times it: 7.227×10237.227\times10^{23}.

Step 1. Number of atoms in 0.4 mol =0.4×NA=0.4×6.022×1023=2.409×1023= 0.4\times N_A = 0.4\times6.022\times10^{23} = 2.409\times10^{23} atoms.

Step 2 (a). In an hcp (close-packed) structure the number of octahedral voids equals the number of atoms: octahedral voids =2.409×1023= 2.409\times10^{23}.

Step 3 (b). The number of tetrahedral voids is twice the number of atoms: tetrahedral voids =2×2.409×1023=4.818×1023= 2\times2.409\times10^{23} = 4.818\times10^{23}. …

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