Density of NaCl is 1.8 g/cc . KCl has similar unit cell as NaCl and the following relationship among ionic radii exist r(Na+)=0.5r(Cl-),r(K+)=0.8r(Cl-) . The density of KCl is

Question

Density of NaCl is 1.8 g/cc . KCl has similar unit cell as NaCl and the following relationship among ionic radii exist r(Na+)=0.5r(Cl-),r(K+)=0.8r(Cl-) . The density of KCl is (A) 2.44 g/cc (B) 1.9 g/cc (C) 1.33 g/cc (D) 0.86 g/cc

Tardigrade Admin · Accepted Answer

Density =(4 × M/NA × a3) (dNaCl/dKCl)=(58 . 5/74 . 5)((aKC l/aNaCl))3 dKCl=1.8× (74 . 5/58 . 5)((1 . 5 r (Cl-)/1 . 8 r (Cl-)))3=1.33 g/cc

Q. Density of $N a Cl$ is $1.8 g / cc$ . $K Cl$ has similar unit cell as $N a Cl$ and the following relationship among ionic radii exist $r (N a^{+}) = 0.5 r (C l^{-}), r (K^{+}) = 0.8 r (C l^{-})$ . The density of $K Cl$ is

$2.44 g / cc$

$1.9 g / cc$

$1.33 g / cc$

$0.86 g / cc$

Density $= \frac{4 \times M}{N _{A} \times a ^{3}}$
$\frac{d _{N a Cl}}{d _{K Cl}} = \frac{58.5}{74.5} (\frac{a _{K Cl}}{a _{N a Cl}})^{3}$
$d_{K Cl} = 1.8 \times \frac{74.5}{58.5} (\frac{1.5 r ( C l ^{-} )}{1.8 r ( C l ^{-} )})^{3} = 1.33 g / cc$

Q. Density of NaCl is 1.8g/cc . KCl has similar unit cell as NaCl and the following relationship among ionic radii exist r(Na+)=0.5r(Cl−),r(K+)=0.8r(Cl−) . The density of KCl is

2.44g/cc

1.9g/cc

1.33g/cc

0.86g/cc