Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

Organic compound → AgBr (s)

  0.5 g                    0.40 g

Here; Moles of Br in organic compound = Moles of Br in Ag Br

= Moles of AgBr

=  $\frac{0.40}{188}mol$

Mass of Br in organic compound =  $\frac{0.4}{188}*80g$  

$\begin{array}{l}\mathrm{\%}ofBr=\frac{0.4*80}{188*0.5}*100\mathrm{\%}\\ =34\mathrm{\%}\end{array}$

All metal carbonyls have synergic bonds

Rate constant, k = 5.5 * 10^-14 s^-1.

$t=\frac{2.303}{k}log\frac{{\left[R\right]}_0}{\left[R\right]}$  

 = $\frac{2.303}{k}log3-(i)$  

$t_{50\mathrm{\%}}=\frac{2.303}{k}log2-(ii)$  

 From (i) & (ii)

$\frac{t_{67\mathrm{\%}}}{t_{50\mathrm{\%}}}=\frac{log3}{log2}$  

= 1.58 t_50%

So;         t_67% is 15.8 * 10^-1 times half life.

X = 16 (the nearest integer)

$C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$  

Here, 6F electricity is required to reduce 1 mol  $C{r}_2{O}_7^{2-}$ to Cr³⁺.

Using ; k_sp = 108 S⁵

1.1 * 10^-23 = 108 S⁵

S =  ${\left(\frac{110}{108}*10^{-25}\right)}^{1/5}M\approx 10^{-5}M$  

Specific conductance,

${\lambda }_m^0=\frac{\kappa }{5*1000}S{m}^{2}mo{l}^{-1}$  

= 3 * 10^-3 Sm² mol^-1

So; x = 3

$P_A^0=50torr$  

$P_B^0=100torr$  

Mole fraction of A in liquid phase,           x_A = 0.3

Mole fraction of B in liquid phase,            x_B = 0.7

Now;     $P_A=P_A^0{x}_A$  

$P_B=P_B^0{x}_B$

= 100 * 0.7 = 70 torr

Mole fraction of B in vapour phase,   $y_B=\frac{P_B}{P_A+P_B}=\frac{70}{85}$  

$\frac{70}{85}=\frac{x}{17}$  

                             X = 14

For expansion in vacuum, workdone, w = 0

For isothermal process,   $\Delta U=0$  

According to first law of thermodynamics,

$\begin{array}{l}\Delta U=q+w\\ q=0\end{array}$  

Hybridization of P in PF₅ is sp³d, so value of y = 1

Work function, ${?}_0=6.63*10^{-19}J$  

Threshold wavelength  ${\lambda }_0=?$  

Using ;                 ${?}_0=\frac{hc}{{\lambda }_0}$  

${\lambda }_0=\frac{hc}{{?}_0}$  

=300 * 10^-9 m = 300 nm

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$

$\begin{array}{l}C{l}_2+2Kl\to 2KCl+I_2\\ I_2+2N{a}_2{S}_2{O}_3\to 2Nal+N{a}_2{S}_4{O}_6\end{array}$

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E}*1000=M*n-factor*V$

$\frac{w}{87/2}*1000=0.1*1*60$

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample = $\frac{0.261}{2}*100$  

 = 13.05%