**a) Molarity**

The formula for the molarity #c# of a solution is

#color(blue)(bar(ul(|color(white)(a/a)c = "Moles"/"Litres" = n/Vcolor(white)(a/a)|)))" "#

Assume that you have 100 g of solution.

Then you have 20 g of ethanol.

#"Moles of ethanol" = 20 color(red)(cancel(color(black)("g EtOH"))) × "1 mol EtOH"/(46.07 color(red)(cancel(color(black)("g EtOH")))) = "0.434 mol EtOH"#

#V = 100 color(red)(cancel(color(black)("g solution")))× "1 mL solution"/(0.996 color(red)(cancel(color(black)("g solution")))) = "100.4 mL solution" = "0.1004 L solution"#

∴ #c = "0.434 mol"/"0.1004 L" = "4.3 mol/L"#

**b) Molality**

The formula for the molality #b# of a solution is

#color(blue)(bar(ul(|color(white)(a/a)b = "moles of solute"/"kilograms of solvent" color(white)(a/a)|)))" "#

Assume that you have 100 g of solution.

Then you have 20 g of ethanol and 80 g of water.

#"Moles of ethanol = 0.434 mol EtOH"#

#"Kilograms of H"_2"O" = 80 color(red)(cancel(color(black)("g H"_2"O"))) × ("1 kg H"_2"O")/(1000 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0800 kg H"_2"O"#

∴ #b = "0.434 mol"/"0.0800 kg" = "5.4 mol/kg"#