Chemistry problem about solutions.

Question

Calculate how many cm^3 of a solution of perchloric acid 55.0% ( density = 1.47 g/mL) must be diluted with water in order to prepare a 0.5 L 0.3 M solution. How much water do we have to add in order to obtain a 0.15 M solution?.

 

I don't manage to solve this problem, I hope someone will help me, thank you so much!

Answer 1

(a) A 0.3 M solution of perchloric acid contains 0.3 mol/L, or .15 mol for .5 L. The molar weight of perchloric acid (HClO₄) is 100.46 g/mol, so we need

100.46 g/mol (.15 mol) = 15.069 g of perchloric acid. With a density of 1.47 g/mL, this corresponds to a pure volume of

15.069 g/(1.47 g/mL) = 10.25 mL = 10.25 cm³.

Our solution is already diluted to 55%, so we need

10.25/0.55 = 18.64 cm³ of the 55% acid.

 

(b) The number of moles, n=VM, stays the same so that V₁M₁ = V₂M₂ , which gives us

0.5(0.3) = V₂(0.15), so V₂= 1 L. You started with 0.5 L, so you need to add another 0.5 L of water to get to 1 L.