0.1 Normal ( or 0.1 Equivalent / L ) = 0.1 Molar ( or 0.1 Mole / L )
For 0.1 Mole of NaOH = 0.1 Mole x ( 40.0 g / Mole of Pure NaOH ) x ( 100 % / 98.5 % ) = 4.0609 g
" Quickly " Weigh 4.07 g of " 98.5 % NaOH - - > Add Distilled / De-Ionized Water to dissolve - - > Adjust volume to 1000 mL - - > If needed - - > Determine the exact concentration By Titration with Potassium hydrogen phthalate ( Volumetric Standard Grade ) .