Asked by sibian
A commercial available sulphuric acid is 15% by mass density=1.10g/mole.calculate molarity and molality of the solution
Answers
Answered by
DrBob222
Your question has an error. The density MUST be 1.10 g/mL, certainly not 1.10 g/mol.
1.10 g/mL x 1000 mL x 0.15 x (1 mol/98 g) = mols/L of solution = ? M
molality = mols H2SO4/kg solvent.
The weight of 1000 mL is 1.10 g/mL x 1000 mL = 1100 grams H2SO4+H2O.
The amount of H2SO4 in that is 1100 x 0.15 = 165 grams.
The amount of H2O is 1100 g - 165 g = 935 g or 0.935 kg.
mols H2SO4 = grams/molar mass = 165/98 = 1.68
m = molality = mols H2SO4/kg solvent = 1.68/0.935 = ?
1.10 g/mL x 1000 mL x 0.15 x (1 mol/98 g) = mols/L of solution = ? M
molality = mols H2SO4/kg solvent.
The weight of 1000 mL is 1.10 g/mL x 1000 mL = 1100 grams H2SO4+H2O.
The amount of H2SO4 in that is 1100 x 0.15 = 165 grams.
The amount of H2O is 1100 g - 165 g = 935 g or 0.935 kg.
mols H2SO4 = grams/molar mass = 165/98 = 1.68
m = molality = mols H2SO4/kg solvent = 1.68/0.935 = ?
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