A 1.0L of an aqueous solution of ethylene glycol, HOCH2CH2OH, is 60% HOCH2CH2OH by mass and has a density of 1.15 g/mL. Calculate the morality of the solution.
2 answers
Sorry I meant molality
molality = moles solute/Kg Solvent
It is a concentration term used mainly for systems that go through a change in temperature and need a constant value for concentration. Concentrations in terms of Molarity would change with changes in temperature due to expansion or contraction of the solution. Anyways...
moles of solute in a 60%EG(aq) soln = (60-g/62-g/mol)/dL (deciliter = 1/10th Liter = 100-ml) Soln = 0.968-mol/dL
Kg Solvent in a 60%EG(aq) soln = (100-60)gms HOH/dL Soln = 40-gms HOH/dL Soln = (40/1000)-Kg HOH/dL soln = 0.04-Kg/dL soln.
molality(m) = moles EG/Kg HOH
= (0.968 mol/dL/0.04 Kg/dL)
= 24.2 molal solution of Ethylene Glycol
It is a concentration term used mainly for systems that go through a change in temperature and need a constant value for concentration. Concentrations in terms of Molarity would change with changes in temperature due to expansion or contraction of the solution. Anyways...
moles of solute in a 60%EG(aq) soln = (60-g/62-g/mol)/dL (deciliter = 1/10th Liter = 100-ml) Soln = 0.968-mol/dL
Kg Solvent in a 60%EG(aq) soln = (100-60)gms HOH/dL Soln = 40-gms HOH/dL Soln = (40/1000)-Kg HOH/dL soln = 0.04-Kg/dL soln.
molality(m) = moles EG/Kg HOH
= (0.968 mol/dL/0.04 Kg/dL)
= 24.2 molal solution of Ethylene Glycol
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