Explanation:
(d) \(20 \,ml\) of \(0.1\,N\) \(NaOH\) neutralize \(20 \,ml \) of \( 0.1\,N\) acid
\({\rm{Weight}}\,\,{\rm{of}}\,\,{\rm{acid}} = 0.126\,\,g\)
Volume \(= 20\, ml=\) \(\frac{{20}}{{1000}}litre\)
Normality $=$ \(0.1 \,N\)
Equivalent weight $= ?$
\({\rm{Equivalent}}\,\,{\rm{weight}} = \frac{{{\rm{weight}}\,\,{\rm{of}}\,\,{\rm{acid}}}}{{N \times V}} = \frac{{0.126 \times 1000}}{{0.1 \times 20}} = 63\)