60. Density of NaCl is 1.8 g/ec.KCI has similar
unit cell as NaCl and the following
relationship among ionic radii exist
r(Na")=0.5 r(CI); r(K)=0.8 r(CI) The
density of KCl is closest to
A) 2.44 g/cc
B) 1.9 g/co
C) 1,33 g/cc
D) 0.86 g/cc
Answers
Given info : Density of NaCl is 1.8 g/ec.KCI has similar unit cell as NaCl and the following
relationship among ionic radii exist ;
r(Na)=0.5 r(CI), r(K)=0.8 r(CI).
To find : the density of KCl is closest to ...
solution : density of unit cell, d = ZM/Nv
where Z is no of atoms per unit cell, M is molecular weight of crystallize compound, N is Avogadro's number and v is volume.
structures of NaCl and KCl are same.
so, Z remains constant. N is already a constant.
therefore, d ∝ M/v ...(1)
edge length of NaCl, a= 2 × distance between Na⁺ and Cl¯ ion
= 2[r(Na⁺) + r(Cl¯)]
= 2[0.5r(Cl¯) + r(Cl¯)]
= 3r(Cl¯)
so, the volume of unit cell, v = a³ = 9 r(Cl¯)²
edge length of KCl, A = 2[r(K⁺) + r(Cl¯)]
= 2[0.8r(Cl¯) + r(Cl¯)]
= 3.6r(Cl¯)
so the volume of unit cell, V = A³ = 12.96 r(Cl¯)²
mass of NaCl = 58.44 amu
mass of KCl = 74.5513 amu
now from equations (1),
density of NaCl/density of KCl = mass of NaCl × volume of KCl/mass of KCl × volume of NaCl
= 58.44 × 12.96 r(Cl¯)²/(74.5513 × 9 r(Cl¯)²]
= 1.13
⇒1.8/density of KCl = 1.13
⇒density of KCl = 1.13 × 1.8 = 2.034 g/cc
Therefore the density of KCl is closest to 1.9 g/cc