Class 12th

Following are the possible isomers of dimethyl cyclopentane.

Following is the reaction for oxidation of oxalic acid by KMnO₄

$2KMn{O}_4+5H_2{C}_2{O}_4+3H_{2}S{O}_4?K_{2}S{O}_4+2MnS{O}_4+10C{O}_2+8H_{2}O$       

Here, Mn²⁺ ion is in product form

Having 5 unpaired electron

Magnetic moment =   BM = 5.916

Cu⁺  Paramagnetic due to unpaired e^-

Cu⁺  Colourless from due to 3d¹⁰

Cu⁺  acts as readucing agent

Cu⁺  it's a product form of Fehling's solution

According to Arrhenius equation ;

$Ink=lnA-\frac{Ea}{RT}$                

Here lnk = 33.24  $-\frac{2.0*10^4}{T}$  

$\therefore \frac{Ea}{R}=2.0*10^4$      

$and{E}_a=2.0*10^4*R=2.0*10^4*8.3=16.6*10^{4}J=166kJ$      

${\lambda }_m^{\infty }$  of Nal = 12.7 mSm² mol^-1

${\lambda }_m^{\infty }$ of NaNO₃ = 12.0 m Sm² mol^-1

${\lambda }_m^{\infty }$ of AgNO₃ = 13.3 mSm² mol^-1

$\therefore {\lambda }_{Agl}^{\infty }=\left[{\lambda }_m^{\infty }\left(AgN{O}_3\right)+{\lambda }_m^{\infty }\left(Nal\right)\right]-{\lambda }_m^{\infty }NaN{O}_3$                

= (13.3 + 12.7) - 12.0 = 14.0 m Sm² mol^-1

  $\Delta T_b=i*K_b*m$

 or $\Delta T_b=K_b*m$   (i = 1 for non-electrolyte)

  $or\frac{{\left(\Delta T_b\right)}_A}{{\left(\Delta T_b\right)}_B}=\frac{{\left(K_b\right)}_A}{{\left(K_b\right)}_B}$

  $or\frac{x}{y}=\frac{1}{8}\left|\therefore y=8\right|$              

In Fe_0.93O

Let us suppose that fraction of Fe²⁺ is x and fraction of Fe³⁺ is (1 – x)

$x*\left(+2\right)+\left(1-x\right)*\left(+3\right)=$ O.S. of Fe atom

$or2x+3-3x=\frac{+200}{93}$

$\therefore \mathrm{\%}ofF{e}^{2+}=85\mathrm{\%}and\mathrm{\%}ofF{e}^{3+}=15\mathrm{\%}$

In Fe_0.93O

Let us suppose that fraction of Fe²⁺ is x and fraction of Fe³⁺ is (1 – x)

$x*\left(+2\right)+\left(1?x\right)*\left(+3\right)=$ O.S. of Fe atom

$or\text{?}\text{?}2x+3?3x=\frac{+200}{93}$

$?\mathrm{\%}\text{?}\text{?}of\text{?}\text{?}F{e}^{2+}=85\mathrm{\%}\text{?}\text{?}and\text{?}\text{?}\mathrm{\%}\text{?}\text{?}of\text{?}\text{?}F{e}^{3+}=15\mathrm{\%}$