Class 12th

Kindly consider the following figure

$R_{f}value=\frac{Distancemovedbysubstancefrombaseline}{Distancemovedbythesolventfrombaseline}=\frac{2}{5}=0.4=4.0*10^{-1}$

$A_2{B}_3?2A^{+3}+3B^{-2}$

123

$\therefore i=1+4\alpha =1+4*0.6=1+2.4=3.4$

$\Delta T_b=i{k}_{b}m=3.4*0.52*1=1.768\approx 1.77K$

$T_b=374.77K\approx 375K.$

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$

$K=\frac{2.303}{t}log\frac{a}{a-x}$

$\frac{kt}{2.303}=log\frac{a}{a-x}$

$\frac{ln2*9}{\frac{10}{3}*2.303}=log\left(\frac{1}{f}\right)$

$log\left(\frac{1}{f}\right)=81.24*10^{-2}=81*10^{-2}$

Gabriel phthalimide synthesis is used for 1° Aliphatic /alicyclic amine

K = 3.3 * 10^-4 s^-1

Time for 40% completion ; t

Using $K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$

3.3 * 10^-4 = $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$  

  $t=\frac{2.303}{3.3}*10^4*0.22\Rightarrow t=25.58mins$

so; the nearest integer is 26.

Coordination no. of an atom in BCC is 8.

  $A^2=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$        

$A^4\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$          

Similarly we get A¹⁹ =   $=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{19}\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\&A^{20}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{20}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

=   $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 1+\alpha +\beta =1gives\alpha +\beta =0........\left(i\right)$        

  $2^{20}+\left(2^{19}-2\right)\alpha =4from\left(i\right)$

$\Rightarrow \alpha =\frac{4-2^{20}}{2^{19}-2}=\frac{4\left(1-2^{18}\right)}{-2\left(1-2^{18}\right)}=-2$          

So, b = 2

Hence b - a = 4

Freundlich adsorption Isotherm

$\left(\frac{x}{m}\right)=k{\left(P\right)}^{\frac{1}{n}}$

At moderate pressure ; $\frac{x}{m}$ $\frac{}{}$ varies non-linearly with P.

So, $x=\frac{1}{n}$

Let the equation of normal is Y – y = - $\frac{1}{m}\left(X-x\right)$  

where m is slope of tangent to the given curve then

$Y-y=-\frac{dx}{dy}\left(X-x\right)$           

It passes through (a, b) so b – y = $\frac{-dx}{dy}\left(a-x\right)$  

->(a – x) dx = (y – b) dy

On integration     $ax-\frac{x^2}{2}=\frac{y^2}{2}-by+c.........\left(i\right)$  

(ii) passes through (3, -3) &  $\left(4,-2\sqrt[]{2}\right)$  then

3a – 3b – c = 9       .(ii)

& 4a -  $2\sqrt[]{2}b$ - c = 12           .(iii)

also given   $a-2\sqrt[]{2}b=3............(iv)$

Solve (ii), (iii) & (iv) b = 0, a = 3

Hence a² + b² + ab = 9