One mole of Sulphuric acid required 100 grams of Caustic soda for complete neutralization. Percentage purity of Caustic soda is

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH} \to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O} ; \\ 1 \mathrm{mole} 2\times 40 \mathrm{g} \\ \mathrm{Weight} \mathrm{of} \mathrm{impure} \mathrm{NaOH} \mathrm{required} =100 \mathrm{g} ; \\ \mathrm{Percentage} \mathrm{purity} \mathrm{of} \mathrm{NaOH} =\frac{80}{100}\times 100=80\% ; \end{gathered}$$