Ncert Solutions Chemistry Class 12th

(a) Shows intra molecular H-bonding.

(b) Shows inter molecular H-bonding.

(c) It does not shows intermolecular H-bonding due to high steric hindrance at o-position of benzene ring.

In the lyophilic colloids, the colloidal particles are extremely solvated.

$O_2\left(g\right)\underset{}{\overset{UV}{\to }}O\left(g\right)+O\left(g\right)$  

$O\left(g\right)+O_2\left(g\right)\stackrel{UV}{\to }{O}_3\left(g\right)$      

Ozone in the stratosphere is a product of UV radiations acting on O₂ molecule.

$2K_{2}C{r}_2{O}_7+8H_{2}S{O}_4+3C_2{H}_{6}O\to 2C{r}_2{\left(S{O}_4\right)}_3+3C_2{H}_4{O}_2+2K_{2}S{O}_4+11H_{2}O$

Given that, rate of appearance of Cr₂(SO₄)₃ = $+\frac{d\left[C{r}_2{\left(S{O}_4\right)}_3\right]}{dt}=2.67mole/min$  

LHS                                                                 RHS

$\left(\frac{Rateofdisappearanceof{C}_2{H}_{6}O}{3}\right)=\left(\frac{RateofappearanceofC{r}_2{\left(S{O}_4\right)}_3}{2}\right)$            

Rate of disappearance of C₂H₆O = $\frac{2.67*3}{2}mol/min$  

Rate of disappearance of C₂H₆O = 4.005 mol /min

Ans. = 4 (nearest integer)

Mass of organic compound = 0.2 gram

$n_{AgBr}=\frac{0.188}{188}mol$

= 0.001 mol

$n_{Br}=n_{AgBr}=0.001mole$

 Mass of Br = 0.001 * 80g

= 0.08 g

Mass & of Br = $\frac{0.08}{0.2}*100\mathrm{\%}$

= 40 %

$2KMn{O}_4\stackrel{\Delta }{\to }\underset{\left(green\right)}{\overset{+6}{K_{2}Mn{O}_4}}+\underset{\left(black\right)}{Mn{O}_2}+O_2$

Mn⁶⁺ has one unpaired electron so paramagnetic and has green colour.

Liquation is applicable for the refining of low melting point metals.

In liquation low melting point metal, melt at first and impurities take more time for melting therefore we can separate metal from their impurities.

FeCl₃ + K₄ [Fe (CN)₆] (Potassium ferrocyanide)

excess

|

K Fe [Fe (CN)₆]

(colloidal species)

Prussian Blue coloured 

colloidal complex

(1) $N{a}^2{O}_2+2H_{2}O\to 2NaOH+H_2{O}_2$  

(2)  $Ba{O}_2.8H_{2}Ogives{H}_2{O}_2$ after evaporation

(3) PbO₂ + 2H₂O ->Pb (OH)₄

(4) SnO₂ + 2H₂O ->Sn (OH)₄