4. You know that one mole of water weights 18 g and its density is 1g/cm 3 . a.

Question

4. You know that one mole of water weights 18 g and its density is 1g/cm3.

a. Calculate density of bulk water (molecules/nm3) and surface density at the air-water interface (molecules/nm2) assuming that average intermolecular distances are the same.       (2)

b. Given that tension at the air-water interface is 72 mN/m (enthalpy of surface formation) and enthalpy of average H-bond is 3.4 kcal/mole, how many H-bonds are dissatisfied at the surface?

                                                                                                                                    (2)

c. If average number of H-bonds in the bulk is 3.5/molecule, what fraction of bonds is dissatisfied at the surface?

Explanation / Answer

4. mass of one mole of water, m = 18 g

density of water, rho = 1g / cm^3

a. assuming average intermolecular distance to be the same

density of bulk water = (1/18) mole / cm^3 = 6.022*10^23/18*(10^7)^3 = 33.51 molecules / nm^3

so 1 nm^3 has 33.51 molecules

1 molecule occupies 1/33.51 nm^3

this is equivalent to a cube of side 0.3101 nm

surface density = 1 molecule / 0.3101^2 nm^2 = 10.3944 molecules per nm^2

b. enthalapy of surface formation, T = 72 mN/m

enthalapy of average H bond, H = 3.4 kCal/mole = 3.4*10^3*4184 J / mole = 2.3622*10^-17 J / molecule

now, nH = TA

where A is surface area per molecule, n is num ber of unsatisfied H bonds, H is average H bond enthalapy

n = TA/H =( 2.3622*10^-17 J / 72*10^-3 N / m * 0.3101^2 * 10^-18 m^2 )^-1

n = 2.93*10^-4 unsatisfied H bonds at the surface

c. fraction of unsatisfied bonds = n/3.5 = 8.3741*10^-5

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.