a) Massof solute = 1.91 g vitamin K
b ) Tf= 6.40 degC
c)Formula;
             Â
Tf=Kf m
Where  Tf is the depression in freezing
point
            Kf
molal depression constant of thesolvent
              mmolality
       m =
Tf/Kf
           =6.40
0C/ 37.8 degC/m
          =0.
169 m
d) Tf=Kf m i
    From that m value found to be 0.169
m.
   molality =no.of mols of solute/1Kg of
solvent
   0.16 mol/Kg = no.of mols of solute/0.025
  no.ofmols of solute =0.169 x 0.025
                                 =0.00423
moles
e) molar mass ofthe solute =mass of solute/no.of mols of
solute
                                        =
1.91 g/0.00423 mols
                                         =451.5
g/mol
