[Vermilion]

What volume of 71% HClO₄ is required to prepare 600mL of a 0.0800M solution in acetic acid?

My attempt: M1V1 = M2V2

                 (0.71)(x) = (0.08M)(0.6L)
                          (x) = [(0.08M)(0.6L)/(0.71)]
                           X = 67.61 mL

I somehow feel that this is incorrect. I cant quite figure out if the percentage of HClO₄ is weight% or volume% or just what it is exactly.  Thanks.

[Borek]

71% w/w is a stock solution, with density around 1.67 g/mL.

[Vermilion]

ahh, okay well, 0.08M x 0.600L = 0.048 mol so

                          ( X₁ )g                     = 0.048 mol so X₁(g) = 4.82208g of HClO4
       (molec weight of HClO4)100.46g

then 4.82208g    =    0.71
             X₂(g)

X₂(g) = 6.792g and then 6.792g times 1mL/1.67g = 4.069 mL I think

[Borek]

4.069 mL I think

Just don't abuse significant digits.