The density of a 1.96 M solution of LiBr in acetonitrile (CH3CN) in 0.826 g/mL. Calculate...
The density of a 1.96 M solution of LiBr in acetonitrile (CH3CN) in 0.826 g/mL. Calculate the concentration of this solution in (a) molality, (b) mole fraction of LiBr, and (c) mass percent of CH3CN.
Homework Answers
Assume the volume of the solution is 1L or 1000 mL
Mass of solution = density x Volume = 0.826 g/mL x 1000 = 826g
Moles of LiBr in the solution = Molarity x Volume (in litres)
= 1.96 x 1 = 1.9mol
Mass of LiBr in the solution = moles x molar mass
= 1.96 mol x 86.9 g/mol = 170.324 g
Mass of acetonitrile. = Total mass - mass of LiBr
= 826 - 170.324 = 655.676 g
Moles of acetonitrile = mass / molar mass = 655.676 / 41 = 15.99 mol
(a) molality = moles solute / kg solvent = 1.96/0.655676 = 3m (approx)
(b) moles fraction of LiBr = moles of LiBr / total moles
= 1.96 / ( 1.96 + 15.99 ) = 0.11
(c) mass percent of CH3CN
= (Mass of acetonitrile/total mass of solution )*100%
= (655.676 / 826)*100 = 79.4%
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