PLEASE HELP!!! PLEASE SHOW ALL STEPS!! THANKS!! 13. The normal boiling point of

Question

PLEASE HELP!!! PLEASE SHOW ALL STEPS!! THANKS!!

13. The normal boiling point of acetone, an important laboratory and industrial solvent, is 56.2 oC and its DHvap is 25.5 kJ/mol. At what temperature (in oC) does acetone have a vapor pressure of 375 torr?

14. Selenium tetrafluoride, SeF4, is a colorless liquid. It has a vapor pressure of 757 mmHg at 105 oC and 522 mmHg at 95 oC. What is its heat of vaporization in kJ/mol?

15. Iron has a body-centered unit cell with a cell dimension of 286.65 pm.
            a) How many iron atoms are in each unit cell?
            b) Calculate the density of iron in g/cm3. (1 pm = 10-12 m)

16. Aluminum metal crystallizes in a cubic close-packed structure (face-centered cubic cell).
            a) How many aluminum atoms are in each unit cell?
            b) Estimate the edge of the unit cell from the atomic radius of Aluminum
                    (1.43 Å, 1 Å = 10-10 m).
            c) Calculate the density of aluminum metal.

Explanation / Answer

13. We will use Calusius-Calpeyron equation,

ln(P1/P2) = -dHv/R[1/T1-1/T2]

where,

P1 = 760 torr

P2 = 375 torr

dHv = 25.5 kJ/mol = 25500 J/mol

T1 = 56.2 oC = 56.2 + 273 = 329.2 K

T2 = unknown

R = gas constant

Feed values,

ln(760/375) = -25500/8.31447[1/329.2-1/T2]

Solving for T2,

T2 = 306.00 K = 33.00 oC

thus, at 33 oC the vapor pressure of acetone will be 375 torr.

14. Again using the Calusius-Calpeyron equation,

ln(P1/P2) = -dHv/R[1/T1-1/T2]

where,

P1 = 522 mmHg

P2 = 757 mmHg

dHv = unknown

T1 = 95 oC = 95 + 273 = 368 K

T2 = 105 oC = 105 + 273 = 378 K

R = gas constant

Feed values,

ln(522/757) = -dHv/8.31447[1/368-1/378]

Solving for dHv,

dHv = 42.989 kJ/mol

Thus, SeF4 has a heat of vaporisation of 42.989 kJ/mol or 43 kJ/mol.

15. Iron has body centered unit cell with cell dimension, 286.65 pm = 2.8665 x 10^-8 cm

a) First calculate volume of unit cell = (2.8665 x 10^-8)^3 = 2.36 x 10^-23 cm^3

So, number of unit cell in 1 cm^3 would be = 1/2.36 x 10^-23 = 4.25 x 10^22 unit cells

b) Calculate the average mass of a 1 atom of Fe [Molar mass = 55.845 g/mol]

= 55.845 / 6.022 x 10^23 = 9.27 x 10^-23 g

So the density of iron = average mass of 1 Fe atom / volume of unit cell

= 9.27 x 10^-23 / 2.36 x 10^-23 = 3.93 g/cm^3

16. Aluminium metal crystallizes in a cubic close packed structure [fcc]

a) fcc cubic close packed has 4 aluminium atoms per unit cell

b) atomic radius = 1.43 Angstrom = 1.43 x 10^-8 cm

across the face of the unit cell, there are 4 radii of aluminium, hence 1.43 x 10^-8 cm.

Using the Pythagorean Theorem, we determine the edge length of the unit cell,

d^2 + d^2 = (5.72 x 10^-8)^2

So, edge length d of unit cell = 4.04 x 10^-8 cm

c) Volume of unit cell = (4.04 x 10^-8)^3 = 6.60 x 10^-23 cm^3

Mass of 4 aluminium atoms = 4 x 27/6.022 x 10^23 = 1.79 x 10^-22 g

So the density of aluminium metal = 1.79 x 10^-22/6.60 x 10^-23 = 2.715 g/cm^3

thus, density of aluminium is 2.715 g/cm^3

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

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