Class 11th

$P_B^O=$ vapour pressure of benzene at 20°C = 70 torr

$P_M^O$ = vapour pressure of toluene at 20°C = 20 torr

Mixture is equimolar, X_B = 0.5 and X_M = 0.5

Total vapour pressure (P_T) = 70 * 0.5 + 20 * 0.5 = 45 torr

Mole fraction of benzene in vapour phase $\left(x_B^{\text{'}}\right)=\frac{P_B^O{X}_B}{P_T}=\left(\frac{70*0.5}{45}\right)torr$

= 0.777 = 77.7 * 10^-2 torr

Ans. = 78 (the nearest integer)

Equation of plane ${\perp }^r$ to line $\frac{x-3}{2}=\frac{x-1}{1}=\frac{x-2}{1}$ and passes through the point

(2, 3, -1) is 2(x – 2) + 1(y – 3) + 1 (z + 1) = 0

=> 2x + y + z – 6 = 0         .(i)

Hence point (1, 2, 2) satisfies equation (i)

$z=\frac{1}{1-cos\theta +2isin\theta }=\frac{1-cos\theta -2isin\theta }{{\left(1-cos\theta \right)}^2-{\left(2isin\theta \right)}^2}=\frac{2si{n}^2\frac{\theta }{2}-2isin\theta }{{\left(1-cos\theta \right)}^2+4si{n}^2\theta }$      

$\therefore Re\left(z\right)=\frac{2si{n}^2\frac{\theta }{2}}{4si{n}^2\frac{\theta }{2}\left(si{n}^2\frac{\theta }{2}+4co{s}^2\frac{\theta }{2}\right)}=\frac{1}{2\left(1+3co{s}^2\frac{\theta }{2}\right)}=\frac{1}{5}\Rightarrow co{s}^2\frac{\theta }{2}=\frac{1}{2}$   

$\Rightarrow \theta =\frac{\pi }{2}\therefore {\displaystyle {\int }_0^{\frac{\pi }{2}}sinxdx=1}$

$6O{H}^{-}+C{l}^{-}\to Cl{O}_3^{-}+3H_{2}O+6e^{-}$

Mole of  $Cl{O}_3^{-}orKCl{O}_3=\frac{10}{122.6}=0.8456mol$

1 mole of $Cl{O}_3^{-}$  or KClO₃ produced by 6 faraday (F).

$\therefore 0.08456$ mole of $Cl{O}_3^{-}$ or KClO₃ produced by (6 * 0.08156) F

$\frac{l*t}{F}=6*0.8156$

$l=\frac{6*0.08156*965000}{10*60*60}A=1.311A\approx 1A$

From   $\Delta H=\Delta G+T\Delta S\therefore \Delta S=\frac{\Delta H-\Delta G}{T}$

$\therefore$ $\Delta S=\left[\frac{51.4-\left(-49.4\right)}{300}\right]*1000J{K}^{-1}mo{l}^{-1}=336J{K}^{-1}mo{l}^{-1}$

Total wt = 4gm = W_NaOH + $W_{N{a}_{2}C{O}_3}$  

Let us suppose moles are 'm' for each.

4 = 40m + 106m

$\therefore m=\left(\frac{4}{146}\right)$ moles of NaOH and Na₂CO₃ each .

$\therefore$ Mass of NaOH (x gm) = $\frac{4}{146}*40=1.095\approx 1.0$

$\left[C{l}^{-}\right]=10^{-1}M$

$\left[Cr{O_4}^{--}\right]=10^{-3}M$

for AgCl ppt¹, ${\left[A{g}^{+}\right]}_{req}=\frac{1.7*10^{-10}}{10^{-1}}M$  

For Ag₂CrO₄ppt², ${\left[A{g}^{+}\right]}_{req}=\sqrt[]{\frac{1.9*10^{-12}}{10^{-3}}}M$

${\left[A{g}^{+}\right]}_{req}=4.3*10^{-5}M$

Being lower concentration of [Ag⁺] in case of AgCl, it will precipitate first.

Regarding industrial consumption of H₂ manufacturing of NH₃ is the largest applications.