You know the concentration of your unknown hcl solution is approximately 0.10 m and you would like to use 40. ml to reach the end point of the titration. what mass of tham should be used to make the 250 ml tham solution?

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.  

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.