of moles of NaOH = (average concentration of NaOH) piersNo. of moles of H2SO4 = 2.3474/2 = 1.1737 molesConcentration of H2SO4 = No. of moles / (volume of diluted acid / 1000) LX 10 mL X 1 L /1000= 1.1737 / (25/1000) (10/1000) L = 0.4694 MTrace 3:Volume of diluted acid = 25 mLVolume of NaOH used = 37.67 mLAverage concentration of NaOH = 0.0881 MNo. of moles of NaOH = (Average concentration of NaOH) of moles of H2SO4 = 2.3387/2 = 1.1693 molesConcentration of H2SO4 = No. of moles / (volume of diluted acid / 1000) LX 10 mL X 1 L /1000= 1.1693 / (25/1000) (10/1000) L = 0.4677 MTrace 4:Volume of diluted acid = 25 mLVolume of NaOH used = 38.32 mLAverage concentration of NaOH = 0.0881 MNo. of moles of NaOH = (Average concentration of NaOH) of moles of H2SO4 = 2.2990/2 = 1.1495 molesConcentration of H2SO4 = No. of moles / (volume of diluted acid / 1000) LX 10 mL X 1 L /1000= 1.1495 / (25/1000) (10/1000) L = 0.4598 MTrace 5:Volume of dilute acid = 25