$H_2O_2$ solution used for hair bleaching is sold as a solution of approximately $5.0 \, g$ $H_2O_2$ per $100 \, mL$ of the solution. The molecular mass of $H_2O_2$ is $34$. The molarity of this solution is approximately $........$ $M$.

An organic compound contains $46.78 \%$ of a halogen $X$. When $2.00 \ g$ of this compound is heated with fuming $HNO_3$ in the presence of $AgNO_3$,$2.21 \ g$ of $AgX$ is formed. The halogen $X$ is:
[Given: atomic weight of $Ag = 108, F = 19, Cl = 35.5, Br = 80, I = 127$]

$A + 2B \rightarrow AB_{2}$. $36.0 \ g$ of '$A$' (Molar mass: $60 \ g \ mol^{-1}$) and $56.0 \ g$ of '$B$' (Molar mass: $80 \ g \ mol^{-1}$) are allowed to react. Which of the following statements are correct?
$(A)$ '$A$' is the limiting reagent
$(B)$ $77.0 \ g$ of $AB_{2}$ is formed
$(C)$ Molar mass of $AB_{2}$ is $220 \ g \ mol^{-1}$
$(D)$ $15.0 \ g$ of '$A$' is left unreacted after the completion of reaction.
Choose the correct answer from the options given below:

Match the following columns:

Column - $A$	Column - $B$
$(A)$ $88 \ g$ of $CO_2$	$(1)$ $0.2 \ mol$
$(B)$ $6.022 \times 10^{23}$ molecules of water	$(2)$ $2 \ mol$
$(C)$ $5.6 \ L$ of $O_2$ at $STP$	$(3)$ $1 \ mol$
$(D)$ $96 \ g$ of $O_2$	$(4)$ $6.022 \times 10^{23}$ molecules
$(E)$ One mole of any gas at $STP$	$(5)$ $3 \ mol$

Dinitrogen and dioxygen react with each other to form different compounds. The experimental data is given below:

Mass of Dinitrogen	Mass of Dioxygen
$(i) \ 14 \ g$	$16 \ g$
$(ii) \ 14 \ g$	$32 \ g$
$(iii) \ 28 \ g$	$32 \ g$
$(iv) \ 28 \ g$	$80 \ g$

$(a)$ Which law of chemical combination is followed by the above experimental data? State the law.
$(b)$ Fill in the blanks in the following conversions:
$(i) \ 1 \ km = \dots \ mm = \dots \ pm$
$(ii) \ 1 \ mg = \dots \ kg = \dots \ ng$
$(iii) \ 1 \ mL = \dots \ L = \dots \ dm^3$

$0.45 \ g$ of an organic compound on combustion gives $0.792 \ g$ of $CO_2$ and $0.324 \ g$ of $H_2O$. In the Kjeldahl method,$0.24 \ g$ of the same compound yields ammonia,which is absorbed in $50 \ mL$ of $0.25 \ N \ H_2SO_4$. The excess acid requires $77.0 \ mL$ of $0.25 \ N \ NaOH$ for neutralization. Determine the empirical formula of the compound. (Atomic masses: $C = 12, H = 1, O = 16, N = 14, S = 32$)