Sucrose (cane sugar), lactose (milk sugar) and maltose are disaccharides in which two monosaccharides are hold together by a glycosidic linkage.

(A) In sucrose, $C_1$ of $\alpha -D$ - glucose and $C_2$ of $\beta -D$ - fructose are linked together (ii).

(B) In lactose, $C_1$ of $\beta$ - D-galactose and $C_4$ of $\beta$-D-glucose are linked together (i).

(C) In maltose, ${\mathrm{C}}_1$ of $\alpha$ - D-glucose is linked to ${\mathrm{C}}_4$ of another $\alpha -\mathrm{D}-$ glucose unit (iii).

So, option (c) is the correct answer.

Aryl amines react with nitrous acid to produce diazonium salts.

Stephen's reduction converts ethane nitrile into ethanal or Acetaldehyde.

When ethane nitrile is treated with Sn/HCl, it undergoes Stephen's reaction to form  Acetaldehyde.

The reaction is:

The reaction is more efficient when aromatic nitriles are used rather than aliphatic nitriles. The reaction is a redox reaction and electron-withdrawing substituents can improve the rate of the reaction.

Heavy water $(D_{2}O)$ freezes at${3.8}^{\circ }C$.

The freezing point of any liquid is the temperature at which the given liquid freezes. It is the temperature for a particular liquid when the liquid turns into a solid state when it undergoes cooling.The temperature for the heavy water at which it will undergo freezing (i.e., it will start to convert itself into the solid-state) is ${3.8}^{\circ }C$ and in the kelvin scale, it is $275.8K$.

${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}+(\mathrm{x}+\frac{\mathrm{y}}{4}){\mathrm{O}}_2\to {\mathrm{xCO}}_2+\frac{\mathrm{y}}{2}{\mathrm{H}}_2\mathrm{O}$

From the reaction,

Produced ${\mathrm{CO}}_2=\mathrm{x}\mathrm{mol}$

and produced ${\mathrm{H}}_2\mathrm{O}=\frac{\mathrm{y}}{2}\mathrm{mol}$

Given produced ${\mathrm{CO}}_2=330\mathrm{g}$

$\therefore$ moles of ${\mathrm{CO}}_2=\frac{330}{44}=\frac{30}{4}=\mathrm{x}$

Also given produced ${\mathrm{H}}_2\mathrm{O}=270\mathrm{gm}$

$\therefore$ Moles of ${\mathrm{H}}_2\mathrm{O}=\frac{270}{18}=15=\frac{\mathrm{y}}{2}$.

$\Rightarrow \mathrm{y}=30$

$\therefore \mathrm{x}:\mathrm{y}=\frac{30}{4}:30=1:4$

Formula of the compound $={\left({\mathrm{CH}}_4\right)}_n$

$\therefore$ Weight of $\mathrm{C}$ in ${\left({\mathrm{CH}}_4\right)}_{\mathrm{n}}=12\mathrm{n}$

Weight of $\mathrm{H}$ in ${\left({\mathrm{CH}}_4\right)}_n=4n$

$\therefore$ Weight ratio of $\mathrm{C}$ and $\mathrm{H}$

$=12\mathrm{n}:4\mathrm{n}$

$=3:1$

$\therefore \mathrm{\%}$ of $\mathrm{C}=\frac{3}{4}\times 100=75$

and $\mathrm{\%}$ of $\mathrm{H}=\frac{1}{4}\times 100=25$

At equilibrium, the relation between the free energy of the reaction and equilibrium constant is as follows,

$\mathrm{\Delta }{G}^{\circ }=-RT\ln K_p$

If $K_p=1$ at equilibrium

$\mathrm{\Delta }{G}^{\prime }=-RT\ln 1$

or $\mathrm{\Delta }{G}^{\circ }=0$

We all know that the hydroxyl group is less reactive than the amine group thus amine group will attack the acetyl chloride group and give acetaminophen commonly called paracetamol.

The reaction is,

As we know that the paracetamol is a drug used to treat torment and fever. It is regularly utilized for mellow to direct torment relief. Evidence is blended for its utilization to assuage fever in children. It is frequently sold in the mix with different meds, for example, in numerous cold medications.

Paracetamol is likewise utilized for serious torment, for example, malignant growth agony and torment after a medical procedure, in the mix with narcotic torment medication. It is ordinarily used either by mouth or rectally, and yet is open by implantation into a vein. Impacts last someplace in the scope of two and four hours.

(D) ${\mathrm{H}}_2\mathrm{S}$

Let oxidation number of sulphur in $H_{2}S$ be $x$.

$\therefore 2+x=0$

$x=-2$

(A) ${\mathrm{H}}_2{\mathrm{SO}}_4$

Let oxidation number of sulphur in ${\mathrm{H}}_2{\mathrm{SO}}_4$ be $x$.

$\therefore 2+x-4(2)=0$

$x=+6$

(B) $S{O}_2$

Let oxidation number of sulphur in $S{O}_2$ be $x$.

$x+(2\times -2)=0$

$x-4=0$

$x=+4$

(C)${\mathrm{H}}_2{\mathrm{SO}}_3$

Let oxidation number of sulphur in ${\mathrm{H}}_2{\mathrm{SO}}_3$ be $x$.

$\therefore 2+x+(3)(-2)=0$

$x=4$