Ineed with these last two questions, !

Both of these questions can be solved using the equation M1V1 = M2V2, where M is concentration anf V is volume.

For the first case, M2 = 0.2 mol/L, M1 = 3 mol/L, and V2 = 250mL. So now you want V1. Solving for V1, V1 = (M2 / M1)V2 =
(0.2 / 3)(250) = 16.7 mL. So what that means is that you need 16.7 mL of 3M HCl, and the rest of the 250 mL (which would be 250 - 16.7 = 233.3 mL) would be water, with which you're diluting the HCl.

Same principle for the second problem, except now we have percentages and not mol/L. You can treat the percentages as concentrations. Since you're starting with pure isopropyl alcohol, M1 = 100%. You want a final volume of 500 mL and a final concentration of 70%. To find the volume of isopropyl alcohol you need to start with, solve for V1. So V1 = (M2 / M1)V2 = (70 / 100)(500) = 350 mL. So you need 350 mL of isopropyl alcohol and the rest of the 500 mL (that is, 150 mL) you can fill with water.

It shows that the experiment is designed properly.