Class 12th

$PC{l}_5\left(g\right)\to PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ (first order reaction)

$K=\frac{2.303}{120}log\left(\frac{50}{10}\right)$      

$\therefore K=\frac{2.303*log5}{120min}=\frac{2.303*0.6989}{120}$ = 0.0134 min^-1

Rate constant at 300K = 1.34 * 10^-2min^-1 [the nearest integer = 1.0]

Total pressure of mixture =   $P_{Total}=X_A*P_A^o+X_B*P_B^o=90*0.6+15*0.4=60mm$

Mole fraction of B in vapour phase, $\left(X_B^{\text{'}}\right)=\frac{X_B*P_B^o}{P_{Total}}$

$\therefore X_B^{\text{'}}=\frac{0.4*15}{60}=0.1=1*10^{-1}$  

In fcc structure of diamond four C present in fcc lattice and other four C present in tetrahedral voids where 50% of tetrahedral voids are occupied. Hence number of carbon atoms present per unit cell of diamond is 8.

$NiC{l}_2+NaCN\underset{O.A.}{\overset{strong}{\to }}{\left[Ni{\left(CN\right)}_6\right]}^{2-}$

Complex has Ni⁴⁺ and strong ligand, hence following are the metal ion electronic configuration

Change of unpaired electron = 2

Wt of Cl^- in 100 ml = 1.8 * 10^-3 gm

Mol. of Cl^- in 100 ml = $\frac{1.8*10^{-3}}{35.5}=0.0507*10^{-3}mole$  

$\therefore \left[C{l}^{-}\right]=\frac{0.0507*10^{-3}*1000}{100}=5.07*10^{-4}M$   

i.e. 0.507 milli mole in one lit required in one hr.

Coagulation value = (millimole/lit) required in one hr = 0.507

= 1 (the nearest integer)

Here, $\lambda =\frac{h}{\sqrt[]{2*m*eV}}=\frac{6.63*10^{-34}}{\sqrt[]{2*1.6*10^{-19}*40000*9.1*10^{-31}}}m$  

$\therefore \lambda =0.614*10^{-11}m$  

$\lambda =6.14*10^{-12}m$ [the nearest integer = 6]

0.15 gm (organic compounds) $\underset{}{\overset{AgN{O}_3}{\to }}AgBr\left(s\right)$

$\downarrow$                                                                    

0.2397 gm

Weight of Br in AgBr = $\left(\frac{80}{188}*0.2397\right)=0.102gm$

% Br in compound = $\frac{0.102*100}{0.15}=68\mathrm{\%}$

Due to the larger acid dissociation constant (Ka) sulphuric acid acts as an acid and HNO₃ acts as a base.

Kindly consider the following figure