D.

Methyl orange is the weak organic base. It is used in the titration of WB vs SA

$\underset{Unionized form Yellow}{MeOH} \leftrightharpoons \underset{Ionized form \left(Red\right)}{M{e}^{+} }+ O{H}^{-}$

In a basic medium, equilibrium lies in the backward direction and therefore it shows yellow colour.

In acidic medium, equilibrium shifts in forwarding direction and therefore, colour changes from yellow to red.

A.

The N-atoms present in the ring will have the same pK_a values (6.0), while N atom outside the ring will have different pKa value (pK_a > 7.4)

Therefore, two N-atoms inside the ring will remain in the unprotonated form in human blood because of their pK_a(6.0) < pH of blood (7.4), while the N-atom outside the ring will remain in protonated form because of its pK_a > pH of blood (7.4).

B.

A and B

$$\begin{gathered} ∆G^{°} = - RT \ln K \\ ∆H° - T∆S° = - RT \ln K \\ -\frac{∆H°}{RT}+\frac{∆S°}{R} = \ln K \end{gathered}$$

Therefore ln K vs 1/T the graph will be a straight line with slope equal to $-\frac{∆H^{°}}{R}$.Since reaction is
exothermic, therefore $∆H^{°}$ itself will be negative resulting in positive slope.

A.

–3267.6

$C_6{H}_6 \left(l\right) + \frac{15}{2} O_2 \left(g\right) \stackrel{ }{\to 6C{O}_2 \left(g\right) +3H_{2}O \left(l\right)}$

Δ_ng = 6 - 7.5
= -1.5 (change in gaseous mole)
ΔU or ΔE = - 3263.9 kJ
ΔH = ΔU + Δn_gRT
Δ_ng = - 1.5
R = 8.314 JK^-1 mol^-1
T = 298 K
So ΔH = -3263.9 + (-1.5) 8.314 x 10^-3 x 298
= -3267.6 kJ
ΔH = Heat at constant pressure
ΔU/ΔE = Heat at constant volume
R = gas constant

A.

C₂H₄O₃

The ratio of mass % of C and H in C_xH_yO_z is 6: 1.
Therefore,
The ratio of mole % of C and H in C_xH_yO_z will be 1: 2.
Therefore x : y = 1 : 2, which is possible in options 1, 2 and 3.
Now oxygen required to burn C_xH_y

$C_x{H}_y + \left(x + \frac{y}{4}\right)O_2 \stackrel{ }{\to } xC{O}_2 + \frac{y}{2}{H}_{2}O$

Now z is half of the oxygen atoms required to burn C_xH_y

$\therefore z = \frac{(2x +{\displaystyle \frac{y}{2}})}{2} = \left(x + \frac{y}{4}\right)$

Now putting values of x and y from the given options:
Option (1), x = 2, y = 4

$$\begin{gathered} z = \left(2 + \frac{4}{4}\right) = 3 \\ Option 2, x = 3, y = 6 \\ z = \left(3 + \frac{6}{4}\right) = 4.5 \\ Therefore correnct \left(C_2{H}_4{O}_3\right) \end{gathered}$$