Question

	Pure Lauric Acid	Lauric Acid + 0.4 g Benzoic Acid	Lauric Acid + 0.5 g Benzoic Acid
Mass of Lauric Acid (g)	3.07	3.05	3.06
Mass of Benzoic Acid (g)	0	0.43	0.51
Freezing Point	43	38	37
Freezing Point Depression, DTf		-5	-6
molality of Benzoic Acid in Lauric Acid		?	?
Moles of Benzoic Acid		?	?
Experimental Molar Mass of Benzoic Acid		?	?
Average Molar Mass		?                              ?
Percent Error		?                              ?

Answer 1

Ans ) Part 1:- For lauric acid and 0.4gbenzoic acid

Here freezing point depression k_f=5

freezing temperature =38

molality of benzoic acid

      =38/5

       =7.9mol/g

Here mass of solvent (lauric acid) =3.05g

Now molality = number of moles of solute (benzoic acid) x 1000/mass of solvent (lauric acid)

7.9 = moles of benzoic acid x1000/3.05

moles of benzoic acid =0.0240

Now mass of solute (benzoic acid) =0.43

mass of solvent =3.05

experimental molecular mass

=

= 5x0.43x1000/38x3.05

=18.55

Average molar mass of benzoic acid C₆H₅COOH = 122g

Percent error = (122-18.55/122) x100

                    = 84.7%

Ans Part 2) For lauric acid +0.5g benzoicacid

Freezing point depression K_f=6

freezing temperature = 37

molality of benzoic acid

      = 37/6

      =6.16mol/g

Here mass of solvent = 3.06

moles of benzoic acid = molality x mass of solvent /1000

                                 =6.16x3.06 /1000

                                 = 0.0188

Now here mass of solute (benzoic acid) =0.51

Experimental molar mass

=

=6x0.51x1000/37x3.06

= 27.02g

Average molar mass of benzoic acid C₆H₅COOH =122

percent error = (122-27.02/122)x100

                   = 77.8%