Question

Calculate the molality of 93% sulphuric acid if the density is 1.84 gram/cc

Answer 1

OK, assuming it is 93% by mass then we should work with 100 g of solution

a 93% by mass solution will have 93 g of H2SO4 and 7 g of H2O

moles H2SO4 = mass / molar mass = 93 g / 98.086 g/mol = 0.948148 moles
mass H2O = 7 g = 0.007 kg

molality = moles solute / kg solvent
= 0.948148 mol / 0.007 kg
= 135 m