Consider the following reactions:
$NaCl + K_{2}Cr_{2}O_{7} + H_{2}SO_{4}$ (conc.) $\rightarrow (A) +$ side products
$(A) + NaOH \rightarrow (B) +$ side product
$(B) + H_{2}SO_{4}$ (dilute) $+ H_{2}O_{2} \rightarrow (C) +$ side product
The sum of the total number of atoms in one molecule each of $(A)$,$(B)$,and $(C)$ is:
$NaCl + K_{2}Cr_{2}O_{7} + H_{2}SO_{4}$ (conc.) $\rightarrow (A) +$ side products
$(A) + NaOH \rightarrow (B) +$ side product
$(B) + H_{2}SO_{4}$ (dilute) $+ H_{2}O_{2} \rightarrow (C) +$ side product
The sum of the total number of atoms in one molecule each of $(A)$,$(B)$,and $(C)$ is:
- A$14$
- B$16$
- C$18$
- D$20$
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