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(B) $1$. In the first reaction, benzene reacts with $6Cl_2$ in the presence of anhydrous $AlCl_3$ (a Lewis acid) under dark and cold conditions. This

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$6$ and $6$

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(B) $1$. In the first reaction, benzene reacts with $6Cl_2$ in the presence of anhydrous $AlCl_3$ (a Lewis acid) under dark and cold conditions. This is an electrophilic substitution reaction where all hydrogen atoms of benzene are replaced by chlorine atoms, resulting in hexachlorobenzene $(C_6Cl_6)$. Thus, $X$ contains $6$ chlorine atoms.
$2$. In the second reaction, benzene reacts with $3Cl_2$ in the presence of $UV$ light at $500 	ext{ K}$. This is a free radical addition reaction. Benzene undergoes addition to form benzene hexachloride $(C_6H_6Cl_6)$, which contains $6$ chlorine atoms. Thus, $Y$ contains $6$ chlorine atoms.
$3$. Therefore, the number of chlorine atoms in $X$ and $Y$ are $6$ and $6$, respectively.

The number of chlorine atoms present in the organic products $X$ and $Y$ of the following reactions, respectively, are:
$\text{Benzene} + 6Cl_2 \xrightarrow[\text{dark, cold}]{\text{Anhydr. } AlCl_3} X$
$\text{Benzene} + 3Cl_2 \xrightarrow{UV, 500 \text{ K}} Y$

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The number of chlorine atoms present in the organic products $X$ and $Y$ of the following reactions, respectively, are: $\text{Benzene} + 6Cl_2 \xrightarrow[\text{dark, cold}]{\text{Anhydr. } AlCl_3} X$ $\text{Benzene} + 3Cl_2 \xrightarrow{UV, 500 \text{ K}} Y$