what is the distance between Na+ and Cl- in NaCl crystal if the desity is 2.165 g /cm cube ? NaCl crystalises in FCC lattic ? At. mass of Na = 23 and Cl = 35.5 a.m.u .please reply fast.
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Distance between Na+andCl−=a2Na+andCl−=a2
Density ρ=2.165gcm−1ρ=2.165gcm−1
Molecular weight of NaCl = 58.5
n = 4 (Since Nacl crystallizes in fcc lattice)
We know that 2(rNa++rCl−)=a2(rNa++rCl−)=a
ρ=nMa3×N×10−30ρ=nMa3×N×10−30
Hence a3=nMρ×N×10−30a3=nMρ×N×10−30
Substituting the value we get
a3=4×58.52.165×6.023×1023×10−30a3=4×58.52.165×6.023×1023×10−30
a=[(4×58.5)2.165×6.023×1023×10−30]13a=[(4×58.5)2.165×6.023×1023×10−30]13
a = 564.17 = 564
2(rNa++rCl−)=5642(rNa++rCl−)=564
(rNa++rCl−)=282pm
Density ρ=2.165gcm−1ρ=2.165gcm−1
Molecular weight of NaCl = 58.5
n = 4 (Since Nacl crystallizes in fcc lattice)
We know that 2(rNa++rCl−)=a2(rNa++rCl−)=a
ρ=nMa3×N×10−30ρ=nMa3×N×10−30
Hence a3=nMρ×N×10−30a3=nMρ×N×10−30
Substituting the value we get
a3=4×58.52.165×6.023×1023×10−30a3=4×58.52.165×6.023×1023×10−30
a=[(4×58.5)2.165×6.023×1023×10−30]13a=[(4×58.5)2.165×6.023×1023×10−30]13
a = 564.17 = 564
2(rNa++rCl−)=5642(rNa++rCl−)=564
(rNa++rCl−)=282pm
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