By titration, 17.6 ml of aqueous h2so4 neutralized 31.4 ml of 0.0111 m lioh solution. what was the molarity of the aqueous acid solution?

The balanced equation for the reaction between LiOH and H₂SO₄ is 
2LiOH + H₂SO₄ → Li₂SO₄ + 2H₂O

Molarity (M) = moles (mol) / volume of the solution (L)

Molarity of LiOH                            = 0.0111 M
Hence, moles of LiOH in 31.4 mL = molarity x volume of the solution
                                                      = 0.0111 M x 31.4 x 10⁻³ L                                                                                               = 3.4854 x 10⁻⁴ mol

The stoichiometric ratio between LiOH and H₂SO₄ is 2 : 1

Hence, reacted moles of H₂SO₄  = moles of LiOH / 2
                                                     = 3.4854 x 10⁻⁴ mol / 2
                                                     = 1.7427 x 10⁻⁴ mol

Those moles were in 17.6 mL of H₂SO₄ solution.
Hence, the molarity of H₂SO₄ = 1.7427 x 10⁻⁴ mol / 17.6 x 10⁻³ L
                                                  = 9.901 x 10⁻³ M

The  molarity of   the aqueous  acid solution  is  9.9   x10^-3  M

   calculation
calculate the moles of LiOH  used

moles =  molarity  x volume in liters
molarity =0.0111 M
volume  in liters = 31.4/1000=  0.0314 liters

moles is therefore = 0.0111  x 0.0314 = 3.485  x10 ^-4  moles  of LiOH

write the equation for neutralization to help to find the moles of H2SO4

= H2SO4 + 2 LiOH →Li2SO4  + 2H2O 

by  use of mole ratio between H2SO4  to LiOH which is  1:2 the moles of  H2SO4 is therefore =  3.485x10^-4  x1/2 = 1.743 x10^-4 moles

molarity   of LiOH=moles/volume in liters

volume in liters = 17.6/1000 =0.0176 liters

molarity  is =  (1 .743 x10^-4) ÷0.0176  =9.9 x10 ^-3 M

The most plausible answer is b, duration. the given amount of time is considered how long it last for.

I’m guessing you’re more alert to yourself or your surroundings if you’re more concentrated