We know the volume of acid×strength of acid = volume of base×strength of base for the neutralization of a acid base titration. Here the volume of acid is 40.0mL and strength of acid is 0.10m (m = molality).
The tham is the tromethamine. The volume of tham in the titration is 250mL.The strength of the tromethamine in the titration with hydrochloric acid (HCl) can be determined as- 40.0×0.10=s(tham)×250. The s(tham) is the strength of the solution which is 0.016m.
Now 1m tham solution is prepared by dissolving 121.14g (as the molecular weight of tham is 121.14g/mole) of tham in 1L of water. Thus to prepare 0.016m solution 121.14×0.016=1.938g of tham have to dissolve in 1000mL of water. Thus in 250mL of tham solution 1.938/4 = 0.484g of tham have to dissolve. Thus the mass of the tham present in the solution is 0.484g.
