Chemistry NCERT Exemplar Solutions Class 11th Chapter Thirteen

${\mathrm{M}\mathrm{a}}_2{\mathrm{b}}_2{\mathrm{c}}_2=5$ Geometrical Isomers

M (AA)b2c2:3  $\mathrm{M}\left(\mathrm{A}\mathrm{A}\right){\mathrm{b}}_2{\mathrm{c}}_2:3$  Geometrical Isomers

Amphiprotic species are those which behave like acid and base both. They can donate and accept a proton.

Kindly consider the following Image 

$\stackrel{\stackrel{+4}{{\mathrm{N}\mathrm{O}}_2}}{2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{N}\mathrm{O}}_2}+\stackrel{+5}{\mathrm{H}\mathrm{N}\mathrm{O}}{}_3$

$\stackrel{0}{{\mathrm{C}\mathrm{l}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{-1}{\mathrm{H}\mathrm{C}\mathrm{l}}+\mathrm{H}\mathrm{O}\mathrm{C}\mathrm{l}$

$2\stackrel{+4}{{\mathrm{C}\mathrm{l}\mathrm{O}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_2}+\stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}$

$\stackrel{+6}{{\mathrm{C}\mathrm{l}}_2}{\mathrm{O}}_6+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}+\stackrel{+{+}^{+}}{{\mathrm{C}\mathrm{l}\mathrm{O}}_4}$

${\mathrm{C}\mathrm{r}\mathrm{O}}_4^{2-}+2{\mathrm{H}}^{+}+2{\mathrm{H}}_2{\mathrm{O}}_2?\underset{\text{Chromium Pentoxide}}{{\mathrm{C}\mathrm{r}\mathrm{O}}_5+3{\mathrm{H}}_2\mathrm{O}}$

$N_1{V}_1=N_2{V}_2$

$0.01*V_1=0.01*20*3$

$V_1=60\mathrm{m}\mathrm{L}$

Molarity $=\frac{\text{moles of}\text{solute}}{\text{volume of solution in}L}=\frac{1*1.15}{1342}*1000=0.86\mathrm{M}$

The number of >C = 0 groups in tripeptide is 0 .

Charge (q)   $=it=2*50*60=6000\mathrm{C}$

$\Rightarrow \mathrm{}6000C=\left(\frac{6000}{96500}F\right)$

${\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}+3{\mathrm{e}}^{(-)}+2{\mathrm{H}}_2\mathrm{O}\stackrel{\text{weakly alkaline medium}}{?}{\mathrm{M}\mathrm{n}\mathrm{O}}_2+4{\mathrm{O}\mathrm{H}}^{-}$

Mass of ${\mathrm{M}\mathrm{n}\mathrm{O}}_2$    deposited $=\frac{1}{3}*\frac{6000}{96500}*87=1.8$

$\to$ ${\pi }_{{\mathrm{K}\mathrm{N}\mathrm{O}}_3}={\pi }_{\text{glucose.}}$

$(\mathrm{i}\mathrm{C}\mathrm{R}\mathrm{T}{)}_{{\mathrm{K}\mathrm{N}\mathrm{O}}_3}=(\mathrm{C}\mathrm{R}\mathrm{T}{)}_{\text{glucose}}$

 $\frac{}{}\frac{}{}$ $\frac{\mathrm{i}*1.34}{101.1}*\mathrm{R}\mathrm{T}=\frac{4.77}{180}*\mathrm{R}*\mathrm{T}$

$\mathrm{i}=\frac{4.77}{180}*\frac{101.1}{1.34}=2$

Hence, extent of dissociation is $100\mathrm{\%}$