Chemistry

Sol. (a) In extraction of iron lime stone is added on a flux

$\begin{array}{rr}& {\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3?\mathrm{C}\mathrm{a}\mathrm{O}+{\mathrm{C}\mathrm{O}}_2\\ & {\mathrm{C}\mathrm{O}}_2+{\mathrm{S}\mathrm{i}\mathrm{O}}_2?{\mathrm{C}\mathrm{a}\mathrm{S}\mathrm{i}\mathrm{O}}_3\end{array}$

Lindlar's catalyst reduces alkynes to cis-alkene.

Lindlar's catalyst reduces alkynes to cis-alkene.

Aromatic species are most stable, here

For ideal gas using ;

PM = dRT

$P=\left(\frac{RT}{M}\right)d$

Comparing with y = mx + C

Slope, m =  $\frac{RT}{M}$

Slope $\propto$  T

So; higher the slope higher the T

Hence, T₃ > T₂ > T₁

Moles of chlorine in the given compound = Moles of chlorine in AgCl

= moles of AgCl

$n_{C\mathcal{l}}=\frac{0.4}{143.5}mo\mathcal{l}$                

Mass of chlorine = $\frac{0.4}{143.5}*35.5g$  

= 0.098 g

$\mathrm{\%}ofC\mathcal{l}=\frac{0.098}{0.25}*100=39.6\mathrm{\%}$        

Using, Freundlich adsorption isotherm ;

$\frac{x}{m}=k.p^{1/n}$       ………………. (i)

$log\frac{x}{m}=\frac{1}{n}logp+logk$                

Comparing with y = mx + C

Slope = $\frac{1}{n}$  = 1

Intercept, log k = 0.602

log k = log 4

k = 4

from equation (i)

  $\frac{x}{m}=4*{\left(0.03\right)}^1$              

= 0.12

= 12 * 10^-2

So, 12 * 10^-2 g of gas is adsorbed per gram of adsorbent,

Initial temperature ; T₁ = 300 K

Final temperature ; T₂ = 309 K

Rate constant gets doubled i.e K₂ = 2K₁

$Usinglo{g}_{10}\frac{k_2}{k_1}=\frac{Ea}{2.3R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$log2=\frac{Ea}{2.3*8.3}\left[\frac{9}{300*309}\right]$

$E_a=\frac{2.3*8.3*300*309*log2}{9}J/mo\mathcal{l}$

$E_a=58.98kJ/mol$    

For the given cell, net redox reaction is as;

$H_2\left(g\right)+2A{g}^{+}\left(aq\right)\to 2H^{+}\left(aq\right)+2Ag\left(s\right)$               

n = 2

$E_{cell}^0=+0.5332V$               

Using;  $\Delta G^0=-nF{E}_{cell}^0$  

$=-2*96487*0.5332J/mo\mathcal{l}$               

$=-102.89kJ/mo\mathcal{l}$               

Now, for the reaction

$\frac{1}{2}{H}_2\left(g\right)+A{g}^{+}\left(aq\right)\text{?}\to H^{+}\left(aq\right)+Ag\left(s\right)$               

n = 1

So; $\Delta G^o=-1*96487*0.5332J/mo\mathcal{l}$  

=  $-51.44kJ/mo\mathcal{l}$

0.001 M NaOH solution has [OH^-] = 0.001 M = 10^-3 M

Using ; pOH = $-lo{g}_{10}\left[O{H}^{-}\right]$  

=  $-lo{g}_{10}10^{-3}$

pOH = 3

pH = 14 – pOH

pH = 14 – 3

pH = 11