6 grams of sodium hydroxide is required to completely neutralize 4.9 grams of sulphuric acid. Percentage purity of sodium hydroxide sample is

$$\begin{gathered} 2\mathrm{NaOH} + {\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O} ; \\ \left(2\times 40\right)\mathrm{g} 98\mathrm{g} \\ 80 \mathrm{grams} \mathrm{of} \mathrm{pure} \mathrm{NaOH} \mathrm{is} \mathrm{required} \mathrm{to} \mathrm{neutralize} 98 \mathrm{grams} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4; \\ \mathbf{4} \mathrm{grams} \mathrm{of} \mathrm{pure} \mathrm{NaOH} \mathrm{is} \mathrm{required} \mathrm{to} \mathrm{neutralize} 4.9 \mathrm{grams} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4; \\ \mathrm{But} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{question},\mathbf{ }\mathbf{6} \mathrm{grams} \mathrm{of} \mathrm{NaOH} \mathrm{was} \mathrm{required}; \\ \mathrm{Percentage} \mathrm{purity} \mathrm{of} \mathrm{NaOH}=\frac{4}{6}\times 100=66.6\% \end{gathered}$$