- A$1$
- B$84$
- C$32$
- ✓$64$
$\%$ of $N$ in the compound $=\frac{25 \times 10^{-3} \times 14 \times 100}{0.55}=63.6$
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$\begin{array}{*{20}{c}}
{C{H_3} - C = C{H_2}C{H_2}OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ is
$\mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{Conc} .) \rightarrow(\mathrm{A})+$ Side products
$(\mathrm{A})+\mathrm{NaOH} \rightarrow(\mathrm{B})+$ side product
$(\mathrm{B})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{dilute})+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow(\mathrm{C})+$ Side product
The sum of the total number of atoms in one molecule each of $(A), (B)$ and $(C)$ is