Step 1. Write the balanced equation
#"2NaOH + H"_2"SO"_4 → "Na"_2"SO"_4 + 2"H"_2"O"#
Step 2. Calculate the equivalents of #"NaOH"#
#"Equivalents" = 0.015 color(red)(cancel(color(black)("L NaOH"))) × "0.1 eq NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.0015 eq NaOH"#
Step 3. Calculate the equivalents of #"H"_2"SO"_4#
In a reaction, 1 eq of anything is equivalent to 1 eq of anything else.
Hence, there are 0.0015 eq of #"H"_2"SO"_4#.
Step 4. Calculate the moles of #"H"_2"SO"_4# in the sample
#"Mass of H"_2"SO"_4 = 0.0015 color(red)(cancel(color(black)("eq H"_2"SO"_4))) × (1 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(2 color(red)(cancel(color(black)("eq H"_2"SO"_4)))) = "0.000 75 mol H"_2"SO"_4#
Step 5. Calculate the mass of #"H"_2"SO"_4# in the sample
#"Mass of H"_2"SO"_4 = "0.000 75" color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("98.08 g H"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.074 g H"_2"SO"_4#
Step 6. Calculate the mass of #"H"_2"SO"_4# in 1 L of solution
#"Mass of H"_2"SO"_4 = 1000 color(red)(cancel(color(black)("mL solution"))) × ("0.074 g H"_2"SO"_4)/(12 color(red)(cancel(color(black)("mL solution")))) = "6.1 g H"_2"SO"_4#
