An aqueous sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution at 20°C has a density of

Question

1.3294 g/mL. Calculate (a) the molarity, (b) the molality, (c) the percent by mass, and (d) the mole fraction of
H2SO4 for the solution. (MM H2SO4 = 98.08; MM H2O = 18.01

I have no idea where to start.

Answer 1

You must remember the definitions. I'll get you started.
M = #moles/liter of solution.
m = # moles/kg solvent.
%w/w = grams H2SO4/100 g solution.
X = moles H2SO4/(moles H2SO4 + moles H2O)

For the first one, molarity.
density = 1.3294 g/mL.
So how much would a liter weigh? It will weigh 1.3294 g/mL x 1000 mL = 1329.4 grams.
How much of that is H2SO4? The problem tells you it has 571.6 g H2SO4 in it.
How many moles is that? It is
571.6 g H2SO4 x (1 mole H2SO4/98.08 g H2SO4) = 5.8279 moles which rounds to 5.828 to 4 significant figures.
So you have 5.828 moles H2SO4/L of solution which makes it 5.828 molar.

I will leave the others to you. Just follow the definitions.

Answer 2

volume of 34.6 and mass of 46.0 what is the SF

Answer 3

The molality is equal to 0.0045146727