Solid State

$Pb{l}_2?P{b}^{2+}+2l^{-}{K}_{sq}=8.0\times 10^{-9}$

Pb (NO₃)₂ -> Pb²⁺ + ^{$2N{O}_3^{-}$}

0.1 M-

-0.1 M    0.1 M

  $K_{sq}=8\times 10^{-9}Using{K}_{sq}=\left[P{b}^{2+}\right]{\left[I^{-}\right]}^2$          

$8\times 10^{-9}=0.1\times {\left(2S\right)}^2$

S = 141 × 10^-6 M

Density = $\frac{M\times z}{N_A{a}^3}$

$d=\frac{63.5\times 4}{6.02\times 10^{23}\times {\left(3.596\times 10^{-10}\right)}^3\times 1000}=9076kg/m^3$

B.C.C structure

a = 300 pm = 300 × 10^-12 m

d = 6g/cm³

z = 2

$d=\frac{Z\times M}{a^3}$

$6=\frac{2\times A}{{\left(300\times 10^{-10}\right)}^3}=\frac{2\times A}{27\times 10^{-24}}$  

$\therefore AtomsofM$ = 3.69 × 6.022 × 10²³

= 22.22 × 10²³

the nearest integer = 22

Density, $\rho =7.62gc{m}^{-3}$  

Cell edge, a = 0.4518nm = 0.458 × 10^-7 cm

Effective number of atoms, Z = 4 for CCP.

Using ; $\rho =\frac{Z\times A}{N_A\times a^3}$  

$\rho =\frac{4\times A}{6.022\times 10^{23}\times {\left(0.458\times 10^{-7}\right)}^3}$              

$7.62=\frac{4\times A}{6.022\times {\left(0.4518\right)}^3\times 10^2}$              

A = 105.79 g/mol

Ans. is 106 (the nearest integers)

Both statement are correct for the glass body heating, but reason is not correct explanation during heating process of glass, constituents unit rupture of glass body and gives the edge smoothness.

In fcc structure of diamond four C present in fcc lattice and other four C present in tetrahedral voids where 50% of tetrahedral voids are occupied. Hence number of carbon atoms present per unit cell of diamond is 8.

Power = 50 watt = 50 J/sec

Energy emitted per second = 50 J

Wavelength ; $\lambda$  = 795 nm

Energy of one photon =   

$=0.025\times 10^{-17}J$             

Number of photons emitted per second = $\frac{50}{0.025\times 10^{-17}}$  

= 2 × 10²⁰

Effective number of atom in C.C.P

$\therefore =\frac{1}{8}\times 8+\frac{1}{2}\times 6$     

= 4

Number of octahedral void = 4

Number of cations = 4

Number of anion = 4

Formula of compound = A₄B₄

Empirical formula = AB

Ans. x = 1

In Frenkel defect, cations are inserted in neighbouring site, therefore statement-I is correct, because vacancy developed by missing cations and interstitial sites are equal. Statement II is false because F-centre developed by metal excess defect due to anion vacancies.

$B{i}_2{S}_3?{\underset{2S}{2Bi}}^{+3}+{\underset{3S}{3S}}^{-2}$

$Ksp={\left(2S\right)}^2{\left(3S\right)}^3=4\times 27\times {\left(S\right)}^5$

= 108 (S)5

${\left(S\right)}^5=\frac{1.08\times 10^{-73}}{108}=10^{-75}$ $\therefore S=10^{-15}M$