A new antibiotic has been isolated but only 200 mg is availa

A new antibiotic has been isolated, but only 2.00 mg is available. The MW is estimated to be 12500. Suppose you wish to check this using colligative properties. Determine the vapor P lowering, freezing point depression, boiling point elevation, and osmotic pressure when the sample is dissolved in 10.0 g water at 20Â °C. [The vapor P of water at 20Â °C is 17.5 torr.]

Solution

Relative lowering of vapour pressure

(P-Ps) = Xb ( P) , P = 17.5 ,

moles of solute = 2x10^ -3/12500 = 1.6 x10^ -7 , moles of water = 10/18 = 0.55 ,

mol fraction of solute Xb = ( 1.6x10^ -7)/(1.6x10^ -7 + 0.55) = 2.909 x10^ -7 ,

now ( 17.5-Ps) = 2.909 x10^ -7 x 17.5

Ps = 17.499994 torr,

Vopor pressure lowering (P-Ps) = (17.5-17.499994) = 6 x10^ -6

Freezing point depression dT = (0-Ts) = kf x m ,

m=molality = ( 1.6 x10^ -7 ) x(1000/10) = 1.6 x10^ -5 ,

dT = 1.86 x 1.6 x10^ -5 = 2.976 x10^ -5 ,

elevation in boiling point

dT = Tb-100 = Kb x m = 0.512 x 1.6 x10^ -5 = 8.192 x10^ -6 ,

Osmotic pressure P = CRT , C = n/V = 1.6x10^ -7 x(1000/10) = 1.6 x10^ -5 ,

P = 1.6 x10^ -5 x 0.0821 x 293 = 3.849 x10^ -4 atm = 0.2925 torr