A compound (moleculr weight 250 g/mol, pure liquid vapr pressure of 2e10 atm at

Question

A compound (moleculr weight 250 g/mol, pure liquid vapr pressure of 2e10 atm at 15 C, and activity coefficient of 0.5) is monitored in the aerosol phase. Overnight (T=15 C), total concentration of this compound in the aerosol phase was measure as 1.5 g/m3. During this time, total organic aerosol particle concentration was measure as 10 g/m3 and the average molecular weight of organic aerosol was 350 g/mol.
a) Based on its partitioning, would you say this compound is volatile or non-volatile at 15 C?
b) What is the equilibrium partitioning coefficient for this compound at 15 C?

Explanation / Answer

a) It is non-volatile
b) Kd = Corg / Cap
   [Corg] = 10 g/m3 * ( (1/250)*10-6 mol / g) = 4*10-8 mol/m3
   [Cap] = 1.5 g/m3 * ( (1/250)*10-6 mol / g) = 6*10-9 mol/m3
Equilibrium Partitioning coeffcient = (4*10-8) / (6*10-9) = 6.667
This shows that the compound is more soluble in organic phase

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