KBr has fcc structure.the density of kbr is 2.75g cm-3 find distance b/w k+ and br-(at mass of br =80.0)
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Formula of density is given by
D = z × M/a³ × Na
Where z is the number of atom per unit cell, M is the molar mass of molecule , a is the edge length and Na is the Avogadro's number.
Now, KBr is fcc {face centered cubic lattice } so, number of atom per unit cell is 4
Hence, z = 4 , density = 2.75 g/cm³ , Na = 6.023 × 10²³ and M = (39+80)= 119 g/mol
∴ D = z × M/a³× Na
2.75 = 4 × 119/a³ × 6.023 × 10²³
⇒a³ = 4 × 119/2.75 × 6.023 × 10²³
⇒a³ = 28.73 × 10⁻²³
after solving it we get , a ≈ 6.59 × 10⁻⁸ cm = 659 pm
Hence, distance between K⁺ and Br⁻ = 659 Pm
D = z × M/a³ × Na
Where z is the number of atom per unit cell, M is the molar mass of molecule , a is the edge length and Na is the Avogadro's number.
Now, KBr is fcc {face centered cubic lattice } so, number of atom per unit cell is 4
Hence, z = 4 , density = 2.75 g/cm³ , Na = 6.023 × 10²³ and M = (39+80)= 119 g/mol
∴ D = z × M/a³× Na
2.75 = 4 × 119/a³ × 6.023 × 10²³
⇒a³ = 4 × 119/2.75 × 6.023 × 10²³
⇒a³ = 28.73 × 10⁻²³
after solving it we get , a ≈ 6.59 × 10⁻⁸ cm = 659 pm
Hence, distance between K⁺ and Br⁻ = 659 Pm
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