How do you start from a #"70%w/w"# #"HClO"_4# solution (density is 1.67) to prepare a #10"%w/v"# solution by starting with a volume of #"5 dm"^3#? #M = 100.46#

I assume your density is #"1.67 g/mL"# and your molar mass is in #"g/mol"#... you didn't supply your units...

Knowing that conc. #"HClO"_4# is #"70%w/w"#, we assume there is

#"70 g HClO"_4# for every #"100 g"# of solution.

[Of course, we could have also said there were #"700 g HClO"_4# for every #"1 kg"# of solution, and it is arbitrary in that sense.]

Knowing the density of the solution, we could then convert the #%"w/w"# to #%"w/v"#, i.e. the #"g"# of solute for every #"mL"# of solution.

#"70 g HClO"_4/(100 cancel"g soln") xx (1.67 cancel"g soln")/"1 mL" xx 100%#

#= 116.9%"w/v"# #"HClO"_4#

Since we want a #10%"w/v"# solution, and we want it to have a volume of #"5 dm"^3#, that is a #116.9:10# dilution, or a dilution factor of #11.69#.

Since the concentration is to decrease by a factor of #11.69#, the volume had to have increased by a factor of #11.69# to reach #"5 dm"^3#.

Therefore, the starting volume would have to been:

#V_1 = ("5 dm"^3)/(11.69) = "0.428 dm"^3 = ul"428 mL"#

The alternative way to do it is to invoke a dilution formula for concentration #c# and volume #V#...

#c_1V_1 = c_2V_2#

#=> V_1 = c_2/c_1V_2#

#= (10%"w/v")/(116.9%"w/v")cdot"5 dm"^3#

#= "0.428 dm"^3 = ul"428 mL"#

as we got with conceptual reasoning.

What I would then do is... (in a fume hood, mind you)

1. Begin with about #"1 dm"^3# of water in a separate #"1-L"# volumetric flask.
2. In a volumetric flask separate from the one the water is in, measure out #"428 mL"# of acid (two #"200-mL"# volumetric flasks, two #"10-mL"# volumetric pipettes, and estimate the last #"8 mL"# in say, a cheap graduated cylinder).
3. In a third giant (#"5-L"# if possible) volumetric flask, pour the water in.
4. In that flask, pour this approx. #"428 mL"# of the acid into the water, and make sure it dissolves. This should be enough headroom to vortex the solution.
5. Dilute it up to the #"5-L"# mark (apparently people make #"5-L"# volumetric flasks).

It is then up to the experimenter to verify the concentration.