Chemistry

23 * 3 gm of Na contained by 164 gm of Na₃PO₄

So, 3.45 gm of Na contained by $\left(\frac{164*3.45}{69}\right)$ gm of Na₃PO₄

Therefore mole of Na₃PO₄= $\frac{164*3.45}{69*164}$ = 0.05 mol

Molarity = $\frac{0.05*1000}{100}$ M = 0.5 M

= 50.0 * 10^-2 M

Stability constant of complex = $\frac{1}{dissocationconstant}$  

$\therefore$ Dissociation constant = $\frac{1}{2.1*10^{13}}=4.76*10^{-14}$  

the nearest value of dissociation constant = 5 * 10^-14

Ans. = 5

  $\underset{Chlorocompound}{\underset{?}{A}}\underset{\left(2\right)H_{2}O}{\overset{\left(1\right)O_3}{\to }}Formaldehyde$

$\downarrow$                   

Weight = 1.53 gm

V = 448 ml (STP)

Mole = $\frac{448}{22400}=\frac{weight}{molecularweight}$

$\therefore \frac{1.53}{molecularweight}=\frac{1}{50}$  

$\therefore$ Molecular weight of compound A = 50 * 1.53 = 76.5 gram

Molecular weight 76.5 = C_nH_2n-1 Cl

$\therefore C_n{H}_{2n-1}=41$  

or 12 n + 2n – 1 = 41

Molecular formula = C₃H₅Cl

Number of C- atoms = 3

Kindly consider the following figure

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.

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

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

= 108 (S)5

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

$E_{cell}={\left(E_{RHS}^o-E_{LHS}^o\right)}_{RP}-\frac{0.0591}{n}lo{g}_{10}Q$

$Q=\frac{\left[Z{n}^{++}\right]}{\left[C{u}^{++}\right]}=\frac{0.04}{0.02}andn=2$

$\therefore E_{cell}=\left(0.34+0.76\right)-\frac{0.0591}{2}lo{g}_{10}\frac{0.04}{0.02}$

$=1.1-\frac{0.0591}{2}*0.3010=1.091V=109*10^{-2}V$

Kindly consider the following figure

${\left[Fe{\left(CN\right)}_{5}NOS\right]}^{4-}$  ${{}_{}}^{}$ Violet colour

${{}_{}}^{}$  ${\left[Fe{\left(SCN\right)}_6\right]}^{4-}$ Blood red colour

$$ $F{e}_4{\left[Fe{\left(CN\right)}_6\right]}_3{H}_{2}O\to$  Prussian blue colour

$$ ${\left[Fe{\left(CN\right)}_6\right]}^{4-}\to$  Yellow colour

${\left[PtC{l}_4\right]}^{2-}+H_{2}O?{\left[Pt\left(H_{2}O\right)C{l}_3\right]}^{-}+C{l}^{-}$  

At equilibrium  $\frac{dx}{dt}=0,$ hence

$0=\frac{-d\left[{\left[PtC{l}_4\right]}^{2-}\right]}{dt}=4.8*10^{-5}\left[{\left[PtC{l}_4\right]}^{2-}\right]-2.4*10^{-3}\left[{\left[Pt\left(H_{2}O\right)C{l}_3\right]}^{-}\right]\left[C{l}^{-}\right]$           

$K_c=\frac{K_f}{K_b}=\frac{4.8*10^{-5}}{2.4*10^{-3}}$              

= 2 * 10^-2

= 0.02