Calculate the molality (m) of a 1.39kg sample of a solution of the solute CH 3 O

Question

Calculate the molality (m) of a 1.39kg sample of a solution of the solute CH3OHdissolved in the solvent C4H8O if the samplecontains 2.67 mol of methanol.
Molar Mass (g/mol) CH3OH 32.04 C4H8O 72.12 Density (g/mL): CH3OH 0.7914 C4H8O 0.8892 Name/Formula: methanol
CH3OH
tetrahydrofuran
C4H8O
molality (m) of a 1.39kg sample of a solution of the solute CH3OHdissolved in the solvent C4H8O if the samplecontains 2.67 mol of methanol.
Molar Mass (g/mol) CH3OH 32.04 C4H8O 72.12 Density (g/mL): CH3OH 0.7914 C4H8O 0.8892 Name/Formula: methanol
CH3OH
tetrahydrofuran
C4H8O

Explanation / Answer

      Molality is the number ofmoles of the solute per kilogram of the solvent .             number of moles of methanol = 2.67              mass of solute = 2.67 * molar mass                                         =2.67 * 32.04g/mol                                        = 85.5 g            massof solution = 1.39 Kg                                     = 1.39 * 103 g                   mass of the solvent = 1304.5 g                molality = 2.67 * 1000 / 1304.5g                                   =2.04 m      

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