Question

A biochemical engineer isolates a bacterial gene fragment and dissolves a 17.3 mg sample of the material in enough water to make 33.5 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?
_________ g/mol

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

__________ degrees C

Answer 1

Osmotic pressure = CRT

Where C = concentration of solution, R =0.0821L.atm/mole.K, T= temperature in K= 25+273.15= 298.15K

Osmotic pressure= 0.34 Torr =0.34/760 atm =0.000447

C= 0.000447/ 0.0821*298.15= 1.83*10-5 moles/L

1 Liter solution contains 1.83*10-5 moles of solute

33.5ml =33.5/1000 L contains 1.83*10^-5*33.5/1000= 6.12*10^-7 moles of solute

moles = mass/molecular weight

Molecular weight= mass/ moles= 17.3*10^-3 gm/ 6.12*10^-7 =28256

b) density of solution= 0.997 g/ml mass of solution =33.5*0.997=33.4 gms

mass of solvent= 33.4-17.3/1000 gm=33.38 gms

33.38 gm   solent contains 6.12*10^-7 moles of solute

1000gm solvent contains 6.12*10^-7*1000/33.38 =1.83*10^-5 moles of solute

this is molality= 1.83*10^-5

Depression in freezing point = kf*m=1.86*1.83*10-5 =3.4*10-5