Question

A biochemical engineer isolates a bacterial gene fragment and dissolves a 13.0 mg sample of the material in enough water to make 39.7 mL of solution. The osmotic pressure of the solution is 0.340 torr at 25°C.

(a) What is the molar mass of the gene fragment?

(b) If the solution density is 0.997 g/mL, how large is the freezing point depression for this solution (K_f of water=1.86°C/m)?

Answer 1

   = iCRT

i = 1 for non electrolyte

   = 0.34 torr

      = 0.34/760   = 0.000447atm

T = 25+273 = 298K

V = 39.7ml   = 0.0397L

W   = 13mg = 13*10^-3 g = 0.013g

   = iCRT

C   = /iRT

      = 0.000447/1*0.0821*298

      = 1.83*10^-5 M

molarity = no of moles/volume in L

1.83*10^-5   = no of moles/0.0397

no of moles = 1.83*10^-5*0.0397   = 7.265*10^-7 moles

molar mass   = weight of substance/no of moles

                       = 0.013/0.0000007265   = 17894g/mole >>>>answer

     

mass of solution = volume * density

                          = 39.7*0.997 = 39.58g

weight of solvent = mass of solution- mass of solvent

                           = 39.58-0.013   = 39.567g

molality = W*1000/G.M.Wt* weight of solvent in g

               = 0.013*1000/17922.75*39.567 = 1.84*10^-5m

Tf = i*m*kf

         = 1*1.84*10^-5*1.86 = 3.42*10^-5⁰C >>>>>>answer

the freezing point depression for this solution = 3.42*10^-5⁰C