Class 12th

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,

 $T=k\sqrt[]{\frac{\rho r^3}{S^{3/2}}}$

$\left[T\right]$ ${\left[M{L}^{-3}\right]}^{1/2}{\left[L^3\right]}^{1/2}{\left[ML{T}^{-2}{L}^{-1}\right]}^{\frac{-3}{4}}$

$\left[T\right]$ $=\left[M^{1/2}{L}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{L}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{M}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{T}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=\left[M^{\frac{1}{4}}{T}^{3/2}\right]$

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

=  $-51.44kJ/mo\mathcal{l}$

Sol. Sol. ⇒  Electric field due to infinite sheet is uniform

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

Mass of solute ; w_B = 2.5 * 10^-3 kg

Mass of solvent, $w_A=75*10^{-3}kg$  

Boiling point of solution ; $T_b=373.535K$  

Boiling point of water ;  $T_b^0=373.15K$

So, elevation in boiling point, $\Delta T_b=T_b-T_b^0$  

$\Delta T_b=0.385K$                                                           

Using ; $\Delta T_b=K_b*\frac{w_B}{M_B*w_A}*1000$  

$0.385=0.52*\frac{2.5*10^{-3}}{M_B*75*10^{-3}}*1000$               

Molar mass of solute ; M_B = 45 g/mol

Number of moles, n = 5 mol

Temperature, T= 300 K

Initial volume, V₁ = 10L

Final volume, V₂ = 20 L

Using;

Work done; w = -2.303 nRT log₁₀   $\frac{V_2}{V_1}$

$=-2.303*5*8*300lo{g}_{10}\frac{20}{10}J$                

= -8630 J

So, magnitude of work done is 8630 J.

Sol. $\stackrel{?}{U}=5\hat{j}$

$\stackrel{?}{a}=10\hat{i}+4\hat{j}$

$\Rightarrow \mathrm{}\stackrel{?}{\mathrm{S}}=\stackrel{?}{\mathrm{U}}\mathrm{t}+\frac{1}{2}\left(\stackrel{?}{\mathrm{a}}\right){\mathrm{t}}^2$

$\Rightarrow \mathrm{}20\hat{i}+y_0\hat{j}=\left(5t^2\right)\hat{i}+\left(5t+2t^2\right)\hat{j}$

$\begin{array}{rr}& 20=5t^2\mathrm{}\mathrm{}{y}_0=5t+2t^2\\ & \mathrm{t}=2\mathrm{}\Rightarrow \mathrm{}18\mathrm{m}\text{.}\end{array}$