% Concentration of Solutions

[SVXX]

The problem is as follows-:
Calculate the volume of 80% by mass of H2SO4, density 1.8 g/mL required to prepare 1L of 0.2 Molar H2SO4.

[Russ]

The problem is as follows-:
Calculate the volume of 80% by mass of H2SO4, density 1.8 g/mL required to prepare 1L of 0.2 Molar H2SO4.

.2M H2SO4 is .2 mol / 1L
Since you're told it's in 1L of solution, we need to calculate the mass of H2SO4 in .2 mol.

n = M/m so therefore m = M/n

m = 98/.2 = 490g

I'm not exactly sure what the 80% by mass refers to (is there .xH2O attached?) but it seems logical that now we multiply by 100/80 to make the percentage up.

490 * 100/80 = 612.5g

Density is 1.8g/mL (in one mL there are 1.8g, in how many mL will there be 612.5g?)

612.5/1.8 = ~340.3 mL

Can somebody advise on what the 80% by mass means? I've never seen a question written that way before.

[SVXX]

As far as I know, 80% by mass is a way of expressing the concentration of solution.

% by mass = (Mass of component in solution)/(Total mass of solution) x 100
This is a binary solution, so we're dealing with only one component in the solvent.
Thus if I take 100 g of solvent, 80 g of H2SO4 is present with 20 g of solvent.
If I divide mass by density, I get the volume for 80% by mass H2SO4.
V = 80/1.8 = 44.44 mL (note that the density is in g/mL).
But hereon there are too many variables and that's what confuses me.