Chemistry 113

Midterm Examination

 

February 8, 2001

 

Name ____________________________

  1. MATCHING: Place the letter corresponding to the word(s) being defined in front of each of the
  2. following (12)

    _C__ An apparatus for measuring atmospheric pressure.

    _R__ The S.I. unit for pressure = 1 N/m2

    _M_ An extrastong dipole-dipole attraction of a hydrogen covalently bonded to a very electronegative atom.

    _I__ The spontaneous intermingling of one substance with another.

    _A__ Description for a noncrystalline solid, i.e. a glass.

    _X__ The smallest portions of a crystal that can be repeated over and over in all directions to give the crystal

    lattice.

    _S__ The conversion of a solid directly into a gas without passing through the liquid state

    _D__ The temperature at which the vapor pressure of a liquid equals atmospheric pressure.

    _Y__ The resistance of a fluid to a change in its form (to flow)

    _V__ The temperature and pressure at which the liquid, solid, and vapor state of a substance can coexist in

    equilibrium

    _L__The enthalpy change associated with gaseous ions or molecules as they dissolve in water.

    _P__ The passage of solvent molecules through a semipermeable membrane

     

     

     

     

    A. Amorphous

    G. Crystal

    M. Hydrogen bond

    T. Surface tension

    B. Atmosphere

    H. Dialysis

    N. London force

    U. Torr

    C. Barometer

    I. Diffusion

    O. Manometer

    V. Triple

    D. Boiling point

    J. Effusion

    P. Osmosis

    W. Vaporization

    E. Colloid

    K. Fusion

    R. Pascal

    X. Unit cell

    F. Critical

    L. Hydration

    S. Sublimation

    Y. Viscosity

  3. MULTIPLE CHOICE: Place the letter(s) corresponding to the best answer(S) on the line in front of

each of the following. (28)

_D___ Of the following gases, the one with the greatest density at STP is

 

A. Ar

B. He

C. Ne

D. Xe

_D___ If a gas’s volume is doubled but the temperature remains constant

 

A. the molecules move faster

B. the molecules move slower

 

C. the pressure is doubled

D. the pressure is halved

_D___ The volume correction term in the van der Waals equation is present because

 

A. manometers are inaccurate

B. molecules are not usually monatomic

 

C. molecules attract each other

D. molecules occupy space

_D___ The boiling points of the rare gases increases going from He to Xe. The type of intermolecular force

that is most responsible for this trend is

 

A. dipole-dipole attraction

B. hydrogen bonding

 

C. ion-ion attraction

D. London dispersion forces

_B___ The intermolecular forces most responsible for the relatively high heat of vaporization for ethanol,

C2H5OH are

 

A. dipole forces

B. hydrogen bonding

C. London forces

D. none of these

_A___ The molecules below all have the same number of electrons and nearly equal molecular weights.

Which one is likely to have the highest vapor pressure?

 

A. CH3-CH2-CH3

B. CH3CH2-F

C. CH3-CH2-OH

D. CH3-O-CH3

_B___ Bromine, Br2, has a triple point at -7.3oC and 44 mmHg and a critical point at 315oC and 102 atm.

The density of the solid is 3.4 g/cm3, and the density of the liquid is 3.1 g/cm3. When cooled

sufficiently at 40 mmHg, bromine vapor

 

A. condenses to liquid

B. condenses to solid

C. does not condense

_A___ Perovskite is a mineral containing calcium, titanium, and oxygen. The unit cell for the mineral has a

Ti ion in the center of the cell, Ca ions at the 8 corners of the cell, and Oxide ions in the 6 faces. The

chemical formula for perovskite is

 

A. CaTiO3

B. CaTiO6

C. Ca8TiO6

D. none of these

_D____ The type of crystal that is soft, low melting, and does not conduct electricity is

 

A. covalent network

B. ionic

C. metallic

D. molecular

_B,(C)_ The most water soluble of the following compounds would be

 

A. C6H6 (l)

B. C2H5OH (l)

C. NaCl (s)

D. SiO2 (s)

_C___ The lattice energy of SrCl2 is (–)2114 kJ/mol. The heat of hydration of SrCl2 (infinite dilution) at

298 K is –2161 kJ/mol. When a very dilute solution of SrCl2 is prepared, the temperature change

expected would be

 

A. none

B. a decrease

C. an increase

_C___ An ideal liquid solution is prepared having equal mole fractions of two volatile components, A

and B. In the vapor above the solution

    1. The mole fraction of A = 0.50
    2. The mole fractions of A and B are equal, but not necessarily 0.50
    3. The mole fractions of A and B are not likely to be equal
    4. There will be only one component present, whichever is the solvent

_B___ The aqueous solution with the lowest freezing point of the following group is

 

A. 0.01 m CH3OH

B. 0.01 m CaCl2

C. 0.01 m CaSO4

D. 0.01 m NaCl

_C___ The best method for determining the molecular weight of a relatively large bio-molecule generally

involves measurement of

 

A. freezing point depression

B. boiling point elevation

C. osmotic pressure

D. vapor density

 

Problems:

  1. An experiment called for 5.83 L sulfur dioxide SO2, at 0oC and 1 atm. What would be the volume of this gas at 25oC and one atm? T up, V up (4)
  2. PV / T = P’V’ / T’ with P constant; 5.83 L / 273 K = v’ / 298 K

    V’ = 5.83 L x 298 / 273 = 6.364 L

    V = |__6.36 L____

     

  3. An 18.6 ml volume of hydrochloric acid reacts completely with a solid sample of Mg CO3. The reaction is 2 HCl(aq) +MgCO3(s) à CO2 (g) + H2O(l) + MgCl2 (aq)
  4. The volume of CO2 formed is 154 ml at 23oC and 731 mmHg. What is the molarity of the HCl solution? PV = nRT; n = PV/RT;

    18.6 mL/1000 mL/L = 0.0186L; 23oC + 273 o = 296 K (8)

    n = [(731 torr/760 torr/atm) x 0.154 L] / [0.08206 Latm/mol deg x 296 K] = 0.0060982 mol CO2

    0.0060982 mol CO2 x 2 mol HCl/1 mol CO2 = 0.01296 mol HCl consumed

    M = mol/L = 0.012196 mol / 0.0186 L = 0.065572 mol/L

    M = |_0.656 M__

  5. If it takes 4.67 times as long for a particular gas to effuse as it takes hydrogen under the same conditions, what is the molecular weight of gas? (5)
  6. Since it takes longer for the unknown gas to effuse, it must be heavier than the hydrogen.

    Time unknown / Time H2 = (MWunk)^0.5 / (MWH2)^0.5 = 4.67 / 1 = (MW)^0.5 / (2.016)^0.5

    (4.67)2 / 12 = MW / 2.016

    MW = 21.8089 x 2.016 = 43.967

    MM = |_ 44.0 g/mol

     

  7. Lithium metal has a body center cubic structure with all atoms at the lattice points and a unit cell length of 350.9 pm. Calculate the density (in g/cm3) of lithium? (8)
  8. Bcc = 1/8 x 8 corners + 1 x 1 center = 2 atoms/unit cell

    Edge = 350.9 pm x 10-12 m/pm x cm/10-2m = 3.509 x 10-8 cm

    D = mass unit cell/ volume unit cell

    = [6.941 g/mol x (1 mol/6.022 x 1023 atoms) x 2 atoms/unit cell]/(3.509 x 10-8 cm)3

    D = 2.305 x 1023 / 4.3207 x 10-23 = 0.53353 g/cm3

    D = |_0.5335 g/cm3

  9. The vapor pressure of benzene is 100.0mmHg at 21.6oC and 400.0 mm Hg at 60.6oC . What is the DHVap for benzene?
  10. 21.6 + 273.2 = 294.8; 60.6 + 273.2 = 333.8 (8)

    ln (p1 / p2) = (DHvap / R) (1/T2 – 1/T1)

    = ln 400.0 torr/100.0 torr = (DHvap /8.314 J/mol deg) (1/294.8 – 1/333.8)

    DHvap = (ln 4.000 x 8.314 J/moldeg) / 3.963 x 10-4 deg-1 = (1.2863 x 8.314 J/moldeg) / 3.963 x 10-4 deg

    = 2.908 x 10-4 J/mol = 29.1 kJ/mol

    DHVap = |_29.1 kJ/mol_

     

  11. Dextran is a polymeric carbohydrate produced by certain bacteria. It is used as a blood substitute. An aqueous solution contains 0.582 g of dextran in 106 ml of solution. It has an osmotic pressure of 1.47 mmHg. What is the average molecular weight of dextran? Assume T = 25oC (8)
  12. 106 mL x 1 L / 1000 mL = 0.106 L; 1.47 torr x 1 atm/760 torr = 0.001934 atm

    P = MRT = (n/V)RT = (mass/MW)/V x RT; MW = mass x RT / V x P

    MW = [0.582 g x 0.08206 Latm/mol deg x 298] / 0.106 L x 0.001934 atm = 6.942 x 104 g/mol

    (if you solve for M instead it = 7.909 x 10-5 mol; n = M x L = 8.384 x 10-6 mol;

    MW = 0.582 g/ 8.384 x 10-6 mol)

    MM = |_6.94 x 104 g/mol

     

  13. An aqueous antifreeze solution is 40.0% ethylene glycol, C2H6O2, by mass The density of the solution is 1.05 g/cm3 Calculate molality, molarity, and mole fraction of ethylene glycol. (15)
  14. MW ethylene glycol = 2(12.01) + 6(1.008) + 2(16.00) = 62.07 g/mol; for H2O = 18.02 g/mol

    The two most straight-forward approaches are :

    Assume 1.00 L of solution; 1.05 g/mL x 1000 mL/ L = 1050g/L solution;

    1050 g x 40.0/ 100.0 = 420 g E.G.; 1050 – 420 = 630 g H2O

    Mol E.G. = 420 g/62.07 g/mol = 6.767 mol; mol H2O = 630 g/ 18.02 g.mol = 34.96 mol

    (1 mL of solution works similarly but with appropriate change of decimal place)

    Assume 100. g soln; 40.0 g E.Gà 40.0 g/ 62.07 g/mol = 0.6444 mol ; 60.0 g H2O à 60.0/18.02 = 3.330 mol

    D = M/V, V = M/D = 100 g solution / 1.05 g/mL = 95.24 mL = 0.09524 L

    (1 g of solution works similarly but with appropriate change of decimal place)

    molality = mol solute / kg solvent = 6.767 mol / 0.630 kg = 0.6444 mol / 0.0600 kg = 10.74 m

    m = |_10.74 molal___

    Molarity = mol solute / L solution = 6.767 mol/L = 0.6444 mol/0.9534 L = 6.766 M

    M = |_6.77 Molar____

    X = mol solute/ mol solution = 6.767 mol /(6.767 + 34.96) mol = 0.644 mol (0.644 mol + 3.330 mol)

    = 0.1622

    X = |_0.162______

     

  15. A solution has a 1:4 mole ratio of pentane to hexane. The vapor pressures of the pure hydrocarbons at 20oC are 441 torr for pentane and 121 torr for hexane. (8)
  1. What are the partial pressures of the two hydrocarbons above the solution?
  2. 1 + 4 = 5 mol total

    PT = XP Ppo + XH PHo = 1/5(441 torr) + 4/5(121 torr) = 88.2 torr + 96.8 torr = 185.0 torr

    Ppentatne = |_88.2 torr___

    Phexane = |_96.8 torr____

  3. What is the mole fraction composition of the vapor?

n proportional to P (ideal gas law); X = pp/PT

XP = 88.2 / 185 = 0.4767

XH = 96.8 / 185 = 0.5332

Xpentane = |_0.477 ____

Xhexanae = |_0.523 ____

Note how the vapor phase is enriched in the more volatile substance.

Bonus. You decide to cool a can of soda pop quickly in the freezer compartment of a refrigerator. When you take out the can, the soda pop is still liquid; but when you open the can, the soda pop immediately freezes. Explain why this happens. (continue on back of page if necessary) (2)

The answer for this bonus must deal with a solution and in particular the freezing point depression of that solution. Two factors would seem important.

The first is that when the can is opened and the gas released from solution, the concentration of the remaining solute is dramatically decreased so that the freezing point of the remaining solution would be raised so that the remaining solution would freeze at a higher temperature than the original.

The second is that when the solute gas escapes from the solution when the can is opened, it must overcome the intermolecular forces it experiences in solution. The escape is thus an endothermic process that will leave the remaining solution colder so that the freezing point is more likely reached.