Equilibrium

$2NOC{l}_{\left(g\right)}?2NO\left(g\right)+C{l}_2\left(g\right)$

t = 0       2 mol/Lit             0            0

t_eq          2 – x                    x            x/2

$K_C=\frac{{\left[NO\right]}^2*\left[C{l}_2\right]}{{\left[NOCl\right]}^2}=\frac{x^2*x}{{\left(2-x\right)}^2*2}=\frac{x^3}{{\left(2-x\right)}^2*2}$      

$=\frac{0.4*0.4*0.4}{1.6*1.6*2}$ = 0.0125 = 125 * 10^-4

(I) $H_2{C}_2{O}_4?H^{+}+H{C}_2{O}_4^{-}$

(II) $H{C}_2{O}_4^{-}?H^{+}+C_2{O}_4^{2-}$

(I) + (II); $H_2{C}_2{O}_4?2H^{+}+C_2{O}_4^{2-}$

$K_{a_3}=K_{a_1}*K_{a_2}$

$\left[N{H}_{4}Cl\right]=\frac{2}{60}=\frac{1}{30}M$

$pH=7-\frac{1}{2}P{K}_b-\frac{1}{2}logC$

$=7-\frac{5}{2}-\frac{1}{2}log\left(\frac{1}{30}\right)=5.24$

Total concentration of H⁺ after the mixing of HCl and H₂SO₄

$\left[H^{+}\right]=\frac{\left(200*0.01\right)+\left(400*2*0.01\right)}{600}$

$\left[H^{+}\right]=\frac{1}{60}$

$\therefore pH=lo{g}_{10}\left(\frac{1}{60}\right)=1.78$

Ka for C₃H₇COOH = 2 * 10^-5

$pKa=-\mathcal{l}og\left(2*10^{-5}\right)=5-\mathcal{l}og2$                

=5 – 0.3 = 4.7

pH of 0.2 (M) solution =

    $pH=\frac{pKa-\mathcal{l}ogC}{2}$            

$=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

$pH=27*10^{-1}$          

$\therefore$ Ans 27

$\underset{\_}{\begin{array}{l}A\left(g\right)?B\left(g\right)+\frac{1}{2}C\left(g\right)\\ t=0amoles00\\ -a\alpha moles+a\alpha moles+\frac{a\alpha }{2}moles\end{array}}$

Eq.   a(1 - α)          aα           (aα/2)

Moles           moles       moles

Total no. of moles at equilibrium

= nA + nB + nC

$=a\left(1-\alpha \right)+a\alpha +\frac{a\alpha }{2}=a\left[1+\frac{\alpha }{2}\right]$

${\left(P_A\right)}_{eq}={\left(x_A\right)}_{eq}*P_{eq}=\frac{a\left(1-\alpha \right)}{\left(1+\alpha /2\right)}P=\frac{1-\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_B\right)}_{eq}={\left(x_B\right)}_{eq}*P_{eq}=\frac{a\alpha }{a\left(1+\alpha /2\right)}P=\frac{\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_C\right)}_{eq}={\left(x_C\right)}_{eq}*P_{eq}=\frac{a\alpha /2}{a\left(1+\alpha /2\right)}P=\frac{\alpha /2}{\left(1+\alpha /2\right)}P$

$K_P=\frac{{\left(P_B\right)}_{eq}*{\left(P_C\right)}_{eq}^{1/2}}{{\left(P_A\right)}_{eq}}=\frac{{\left(\alpha \right)}^{3/2}{P}^{1/2}}{\left(1-\alpha \right){\left(2+\alpha \right)}^{1/2}}$

Metallic suphides i.e CdS are negatively charged colloidal solution.

Starch colloid is macro molecular colloidal solution.

Fe₂O₃. xH₂O sol is positively charged colloidal solution.

Cheese is an example of gel.

Data is not sufficient but official answer is (D)

Aromaticity drives the highest enolic percentage of given structure:

During removal of temporary hardness of water.

$\begin{array}{l}Mg{\left(HC{O}_3\right)}_2\stackrel{Boil}{\to }Mg{\left(OH\right)}_2+2C{O}_2\uparrow \\ Ca{\left(HC{O}_3\right)}_2\stackrel{Boil}{\to }CaC{O}_3+H_{2}O+C{O}_2\end{array}$