$1 \ g$ of a carbonate $(M_2CO_3)$ on treatment with excess $HCl$ produces $0.01 \ mol$ of $CO_2$. The molar mass of $M_2CO_3$ is $.......... \ g \ mol^{-1}$. (Nearest integer)

Match the following items in List-$I$ with the corresponding results in List-$II$ (at $STP$):

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

$10 \ mL$ of conc. $H_2SO_4$ $(18 \ M)$ is diluted to $1 \ L$. The approximate strength of the dilute acid is $........... \ N$.

To $50 \ mL$ of $0.1 \ N \ Na_2CO_3$ solution,$150 \ mL$ of water is added. What is the molarity of the resultant solution?

The number of molecules of $CO_2$ liberated by the complete combustion of $0.1 \ mol$ of graphite in air is

An unknown chlorohydrocarbon has $3.55\%$ of chlorine. If each molecule of the hydrocarbon has one chlorine atom only,the number of chlorine atoms present in $1\,g$ of chlorohydrocarbon is (Atomic wt. of $Cl = 35.5\,u$; Avogadro constant $= 6.023 \times 10^{23}\,mol^{-1}$)