b
\(\%\) of \(N=\frac{1.4 \times \text { meq.of acid }}{\text { mass of organic compound }}\)

meq. of \(H_{2} S O_{4}=60 \times \frac{M}{10} \times 2=12\)

meq. of \(N a O H=20 \times \frac{M}{10}=2\)

meq. of acid consumed \(=12-2=10\)

\(\therefore \% \quad N=\frac{1.4 \times 10}{1.4}=10 \%\)