Calculate the molality of a sulfuric acid solution of specific gravity 1.2 containing 27% H2SO4 by weight.
THE ANSWER IS 3.8 FOR YOU REFERENCE.
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27% H2SO4 by mass means 1000g of solution contains 270g of H2SO4
Number of moles of H2SO4 = 270g / (98 g/mol) = 2.75 mol
Mass of solvent = 1000g - 270g = 730g = 0.73 kg
Molality = Number of moles of solutes / Mass of solvent in kg
Molality = 2.75 / 0.73 ≈ 3.8m
Number of moles of H2SO4 = 270g / (98 g/mol) = 2.75 mol
Mass of solvent = 1000g - 270g = 730g = 0.73 kg
Molality = Number of moles of solutes / Mass of solvent in kg
Molality = 2.75 / 0.73 ≈ 3.8m
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Answer:
3.8m
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