Mass of B which is combined with fixed mass of A (say 1 g) will be 2.5 g, 5 g, 1.25 g and 3.75 g in AB,  AB₂, A₂B and  A₂B₃  respectively. They are in the ratio 2 : 4 : 1 : 3. which is simple whole number ratio. Hence, the law of multiple proportion is applicable.

(i)  Yes
(ii)  According to the law of multiple proportions.
(iii) \( \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \\ 2 \mathrm{~g} \quad 16 \mathrm{~g} \quad 18 \mathrm{~g} \)
\(\mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2} \\ 2 \mathrm{~g} \quad 32 \mathrm{~g} \quad34 \mathrm{~g} \)
Here, masses of oxygen, (i.e. 16 g in H₂ O and 32 g in H₂ O₂) which combine with fixed mass of hydrogen (2 g) are in the simple ratio i.e. 16 : 32  or 1 : 2.

To proceed the calculation, first, calculate the number of moles of NaOH and H₂O
Then, find mole fraction of NaOH and H₂O by using the 
formula \({ { X }_{ NaOH }= }\frac { \eta _{ NaOH } }{ \eta _{ NaOH }+{ \eta }_{ { H }_{ 2 }O } } \left( orX_{ { H }_{ 2 }O }\frac { { \eta }_{ { H }_{ 2 }O } }{ \eta _{ NaOH }+{ \eta }_{ { H }_{ 2 }O } } \right) \)
Then calculate molarity \(=\frac { W\times 1000 }{ m\times V } \) so in order to calculate molarity we reqiure volume of solution which is, \(V=\frac { m }{ specific \ gravity } \)
Number of moles of NaOH,  \(\eta _{ NaOH }=\frac { 4 }{ 40 } =0.1mol \ \left\{ \because n=\frac { Mass(g) }{ Molar \ mass(gmol^{ -1 }) } \right\} \)
Similarly, \(\eta _{ { H }_{ 2 }O }=\frac { 36 }{ 18 } =2 \ mol\)
Mole fraction of NaOH,
\(X_{ NaOH }=\frac { moles \ of \ NaOh }{ moles \ of \ NaOH+moles \ of \ { H }_{ 2 }O }\)
\( X_{ NaOH }=\frac { 0.1 }{ 0.1+2 } =0.0476\)
Similarly, \(X_{ { H }_{ 2 }O }=\frac { { \eta }_{ H_{ 2 }O } }{ { \eta }_{ NaOH }+{ \eta }_{ { H }_{ 2 }O } } =\frac { 2 }{ 0.1+2 } =0.9524\)
Total ,mass of solution = mass of solute + mass of solvent
= 4 + 36 = 40 g
Volume of solution
\(=\frac { mass \ of \ solution }{ specific \ gravity } =\frac { 40g }{ 1 \ gm{ L }^{ -1 } } =40 \ mL\)
Similarly,\(=\frac { mole \ of \ solute\times 1000 }{ volume \ of \ solution(mL) } =\frac { 0.1\times 1000 }{ 40 } =2.5M\)

 \( 2 \mathrm{~N}_{2}(g) +\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{~V} \quad \quad\quad 1 \mathrm{~V} \quad \quad \quad 2 \mathrm{~V} \)
 45.4 L  \(\longrightarrow \)  22.7 L  \(\longrightarrow \)  45.4 L
\(\frac { 45.4 }{ 22.7 } = 2\)      \(\frac { 22.7 }{ 22.7 } = 1\)  \(\frac { 45.4 }{ 22.7 } = 2\)
Hence, the ratio between the volume of the rectants and the product in the given question is simple
i.e. 2: 1 : 2. It proves the Gay Lussac's law of gaseous volumes.
For Gay Lussac's law of gaseous volumes.

There are three common scales to measure temperature ^oC (degree celsius), ^oF(degree Fahrenheit) and K (kelvin). The K is the SI unit.
The temperature on two scales are related to each other by the following relationship \(^{ 0 }F=\frac { 9 }{ 5 } t^{ 0 }C+32\) 
Putting the values in above equation.
\(200-32=\frac { 9 }{ 5 } t^{ 0 }C\Rightarrow \frac { 9 }{ 5 } t^{ 0 }C=168\)   
\(\Rightarrow \ t^{ 0 }C=\frac { 168\times 5 }{ 9 } =93.3^{ 0 }C\)