Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

${\Delta }_t=\frac{hc}{\lambda }=\frac{6.63*10^{-34}*3.08*10^8}{600*10^{-9}}$

$=\frac{6.63*3.08*10^{-17}}{600}$ J

$=765.765*10^{-21}J$

$?766*10^{-21}$ J

$\frac{d\left[C\right]}{dt}=\left(\frac{20-10}{10}\right)=1mmold{m}^{-3}$

$+\frac{d\left[D\right]}{dt}=\left\{-\frac{d\left(B\right)}{dt}\right\}*1.5=\frac{3}{2}\left\{-\frac{d\left[B\right]}{dt}\right\}$

$\left\{-\frac{d\left[B\right]}{dt}\right\}=2*\left\{\frac{-d\left[A\right]}{dt}\right\}$

$\therefore \frac{1}{6}\left\{\frac{-d\left[B\right]}{dt}\right\}=\frac{1}{3}\left\{\frac{-d\left[A\right]}{dt}\right\}=\frac{1}{9}\left\{\frac{+d\left[D\right]}{dt}\right\}=\frac{d\left[C\right]}{dt}$

$\therefore$ rate of reaction = $\frac{+d\left[C\right]}{dt}$  = 1m.m dm^-3 S^-1

$F{e}^{+3}+l^{-}\to I_2+F{e}^{+2}$

Fe⁺² undergoes reduction & I₂ undergoes oxidation.

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

= (0.77 – 0.54) V = 0.23 V

 = 23 * 10^-2 V

$\therefore$  x = 23

 x = 23

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2*0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}*1\right)}^2}{\left(\frac{0.5}{1.5}*1\right)}$

$=\frac{4}{3}=1.33$

$=-710.15J/mol$

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

                             = 0.00002 * 10³ mm

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165*1000}{-550}K$

$\begin{array}{l}=3*100K\\ =300K\end{array}$

$E_{1_H}=-2.2*10^{-18}J$

$Li\to L{i}^{+2}+2e^{-},n=2$

$E_{L{i}^{+2}}=E_{1H}*\frac{Z^2}{n^2}=-2.2*10^{-18}*\frac{3^2}{2^2}$

$\lambda =4*10^{-8}$ m

$d=\frac{ZM}{N_A{a}^3}=\frac{4*58.5}{6.02*10^{23}*a^3}$

$43.1=\frac{4*58.5}{6.02*10^{23}*a^3}$

$\therefore a=2.08*10^{-8}cm$

$\therefore r_{N{a}^{+}}+r_{C{l}^{-}}=1.04*10^{-10}$ m

Molar mass of C₇H₅N₃O₆ = 12 * 7 + 1 * 5 + 14 * 3 + 16 * 6 = 84 + 42 + 96 = 227 g/mol

Number of moles =  $\frac{681}{227}$  = 3 mol

$\therefore$  number of N-atoms

$=3*6.02*10^{23}*3=9*6.02*10^{23}$  

$=5418*10^{21}$  

$\therefore x=5418$