Question

A biochemical engineer isolates a bacterial gene fragment and dissolves a 17.0 mg sample of the material in enough water to make 38.9 mL of solution. The osmotic pressure of the solution is 0.340 torr at 25

Answer 1

Hi,

A) ? = i M RT

? = osmotic pressure = 0.340 torr x (1 atm / 760 torr) = 4.47x10^-4 atm
i = van't hoff factor = # ions the solute dissocates into = 1 in this case
M = molarity = moles solute / L solution
T = temperature = 298 K

before we do any other calculations I'm going to rearrange the above equation...

? = i M R T
M = moles / L solution
moles = mass / mw

substituting
M = (mass / mw) / L solution

substituting
? = i x [ (mass / mw) / L solution ] RT

rearranging
mw = i x mass x R x T / (? x L solution)

ok....

mw = (1) x (0.0170 g) x (0.08206 Latm/moleK) x (298K) / [(4.47x10^-4 atm) x (0.0389 L)] = 2.39x10^4 g/mole

Molar mass of the gene fragment is 2.39x10^4 g/mole

B) ?T = Kf x m x i

?T = change in freezing point
Kf = 1.86 C / m
m = molality = moles / kg solvent
i = 1

let's find m...

38.9 mL x (0.997 g / mL) = 38.7833 g solution.

mass water = mass solution - mass fragment = 38.7833g - 0.0170 g = 38.7663 g water = 0.03877 kg water

so m = (0.0170 g) x (1 mole / 2.39x10^4 g) / 0.03877 kg = 1.83x10^-5 molal

so ....

?T = Kf x m x i = (1.86 C / m) x (1.83x10^-5 m) x 1 = 3.40 x10^-5 C

The freezing point depression for this solution is 3.40 x10^-5 C