Question

Consider a 3.5m solution of glycerol (C3H8O3). density =1.0585 g/mL. a) what is the percentage glycerol by weight? b) what is the molarity of the solution? c) what is the mole fraction of the solution?

Answer 1

Answers,

........a)

.................Explanation & calculations

(i)...................The Molality (m) of a solution is given by the relation ,

...........................................................................................m = No. of moles of solute / Mass` of solvent in kg

..So, applying the above relation for molality of glycerol , 3.5 = No. of moles of glcerol per 1000 gms of solvent

  3 moles of glycerol (C3H8O3 , mol. mass 92 ) = ( 3.5 x 92 ) = 322 gms of glycerol are present in 1000 gms. of water ( ie.solvent )

(ii) Weight percent of a solution is given by the relation ,

...........................Weight per cent = ( weight of solute / weight of solution ) x 100

...........................-------------------= [ 322 gms. / ( 1000 + 322 ) gms.] x 100

......................................................= [ (322 / 1322)x 100 ]

.......................................................= 24.357 %

b)

........Given , density of the solution = 1.0585 g/ml

........so (i) calculate volume of the solution with a total weight = 1322 gms . ......(.Calculated as above )

....................................................................volume of solution = Mass / Density

....................................................................................................= 1322 gms. / 1.0585 gms.per ml

.....................................................................................................= 1248.94 ml

......................................................................................................= 1.248 L

(ii) ................................................................ Now, Molarity (M) = No. of moles of solute / Volume of solution in litre

substituting the given number of moles of glycerol as 3.5 we get,

...................................................................................Molarity (M ) = 3.5 / 1.248

........................................................................................................= 2.8044 M

............c).Calculation for mole fraction

.........................(i)...Mole fraction of solute = moles of solute / ( Moles of solute + Moles of solvent )

......................... (ii) Moles of solvent = ( 1000 / 18)

.....................................................................= 55.55 moles

..........................(iii) substituting the values we get,

.................mole fraction of glycerol in solution = [ 3.5 / ( 3.5 + 55.55 ) ] moles   

..........................................................................= ( 3.5 / 59.05 ) moles

...........................................................................=0.0593 moles

.Mole fraction of solvent (water ) in solution = ( 1.00 - 0.0593) moles

.............................................................................= 0.9407 moles