Question Solved1 Answer 1) Using the molecular weights and densities calculate the volumes required to prepare a 1:1 mole ratio solution of acetone and toluene. da =0.791g/mL MWA =58.19/mol 51gmot * O.791g/mL x Imol = 0. 013 6 mol C=n 58 g %3D

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Transcribed Image Text: 1) Using the molecular weights and densities calculate the volumes required to prepare a 1:1 mole ratio solution of acetone and toluene. da =0.791g/mL MWA =58.19/mol 51gmot * O.791g/mL x Imol = 0. 013 6 mol C=n 58 g %3D
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Transcribed Image Text: 1) Using the molecular weights and densities calculate the volumes required to prepare a 1:1 mole ratio solution of acetone and toluene. da =0.791g/mL MWA =58.19/mol 51gmot * O.791g/mL x Imol = 0. 013 6 mol C=n 58 g %3D
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Answer-Density of acetone = 0.791 g/mLDensity of toluene = 0.87 g/mLmolar mass of acetone = 58.08g/molmolar mass of toluene = 92.14g/molmole ratio = 1:1Volumes required = ?We know that,mole ratio = 1:1i.e. 1 mole of Acetone and 1mole of TolueneAlso,Moles = Mass/Molar MassMass = Moles * MolarMassSo, mass of 1 mole of Acetone = 1 mol* 58.08 g/mol = ... See the full answer