The equation for the reaction is
#"H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + 2"H"_2"O"#.
1. Calculate the moles of #"H"_2"SO"_4#
#"Moles of H"_2"SO"_4 = "0.035 21" color(red)(cancel(color(black)("L H"_2"SO"_4))) ×( "0.550 mol H"_2"SO"_4)/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) = "0. 019 37 mol H"_2"SO"_4#
2. Calculate the moles of #"NaOH"#
#"Moles of NaOH" = "0.019 37" color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("2 mol NaOH")/(1 color(red)(cancel(color(black)("H"_2"SO"_4)))) = "0.038 73 mol NaOH"#
3. Calculate the molarity of the #"NaOH"#
#"Molarity" = "moles"/"litres" = "0.038 73 mol"/("0.050 00 L") = "0.775 mol/L"#
