Quiz 3 - Properties of Liquids | Liquids and Solids

General Chemistry 2 - Quiz 3 - Properties of Liquids

1

The substances below have molar masses that are the same within  2 g/mol. Which substance has the lowest boiling point?

Propane (CH3CH2CH3) has the lowest boiling point because it is nonpolar and only experiences weak London dispersion forces. In contrast, the other substances have stronger intermolecular forces, such as dipole-dipole interactions or hydrogen bonding, leading to higher boiling points.

2

Which statement is correct about the critical point of a phase diagram?

At the critical point on a phase diagram, the liquid and vapor phases become indistinguishable from each other. Beyond this point, the substance exists as a supercritical fluid, where the distinctions between the liquid and gas phases disappear.

3

When equal volumes of the following pairs of liquids are mixed thoroughly and allowed to stand, which pair is most likely to separate into two layers?

Carbon tetrachloride (CCl4) is a nonpolar molecule, while methanol (CH3OH) is a polar molecule with hydrogen bonding. Because polar and nonpolar substances do not mix well, CCl4 and methanol are likely to separate into two layers when mixed.

In contrast, ethanol and methanol are both polar and miscible, hexane and pentane are both nonpolar and miscible, and carbon tetrachloride and hexane are both nonpolar and miscible.

4

Which physical property decreases with an increase in intermolecular forces?

As intermolecular forces increase, the molecules in a liquid are held together more strongly, making it harder for them to escape into the vapor phase. This results in a decrease in vapor pressure. In contrast, the boiling point, enthalpy of vaporization, and viscosity all increase with stronger intermolecular forces.

5

The atmospheric pressure on the summit of Mt. Everest is 0.333 atmospheres. At what temperature (in oC) does H2O boil there? (ΔHvap H2O = 40.7 kJ.mol–1)

To find the boiling point of water at the summit of Mt. Everest, where the atmospheric pressure is 0.333 atmospheres, we can use the Clausius-Clapeyron equation:

 

 

ln P2P1 = - HvapR 1T2 - 1T1

P1 = 1 atm (standard boiling point of water at 100 oC)
P2 = 0.333 atm (pressure at the summit of Mt. Everest)
ΔHvap​ = 40.7 kJ.mol-1 = 40700 J.mol-1
R = gas constant = 8.314 J.mol-1.K-1
T1 = 373 K (100 oC)
T2 = ? (boiling point temperature at 0.333 atm in K)

 

Therefore,

ln 0.3331 = - 407008.314 1T2 - 1373

⇒  - 1.100 = - 4895 1T2 - 1373

⇒  2.247 x 10-4 = 1T2 - 1373

⇒ 1T2 = 2.906 x 10-3

⇒ T= 344 K = 71 oC

6

Which of the following substances experience London dispersion forces?
I. CH3CH3         II. CH3OH

London dispersion forces are present in all molecules, whether polar or nonpolar. They arise due to temporary fluctuations in the electron distribution within molecules, which induce temporary dipoles.

  • CH3CH3 is a nonpolar molecule, so its intermolecular forces are primarily London dispersion forces.
  • CH3OH is a polar molecule and primarily experiences hydrogen bonding. However, it also experiences London dispersion forces in addition to hydrogen bonding.

Therefore, both substances experience London dispersion forces.

7

Which of these pure substances has the highest normal boiling point?

Boiling point is influenced by the strength of intermolecular forces:

  • CH4 experiences only London dispersion forces, which are weak.
  • SiH4 also experiences only London dispersion forces, similar to CH4.
  • PHhas weak dipole-dipole interactions, but they are not as strong as hydrogen bonding.
  • NHhas strong hydrogen bonding due to the presence of nitrogen bonded to hydrogen, which significantly increases its boiling point.

Hydrogen bonding is much stronger than the other intermolecular forces present in CH4, SiH4, and PH3, leading to NH3 having the highest normal boiling point among the listed substances.

8

Nitrogen, N2, has the following properties:
normal melting point: 63.2 K; normal boiling point: 77.4 K; triple point: 0.127 atm, 63.1 K; critical point: 33.5 atm, 126.0 K
Which statement about N2 is correct?

  • Option A is incorrect because, typically, solids are denser than liquids for most substances, including nitrogen (except in anomalous cases like water). Solid nitrogen is denser than liquid nitrogen.
  • Option B is incorrect because nitrogen cannot be liquefied at 150 K since its critical temperature is 126.0 K. Above this temperature, nitrogen cannot be liquefied regardless of the pressure.
  • Option C is incorrect because at 63.1 K and 1 atm, nitrogen would be in the solid state, not a liquid-gas equilibrium. The triple point occurs at 63.1 K and 0.127 atm, where solid, liquid, and gas phases coexist.
  • Option D is correct. Below the triple point (63.1 K, 0.127 atm), nitrogen sublimates, meaning it transitions directly from solid to gas without passing through the liquid phase. At 0.100 atm, which is below the triple point pressure, nitrogen will sublime when heated from 60 K to 70 K.
9

Which characteristic of this curve is related to the value for the enthalpy of fusion of the substance?

The enthalpy of fusion (ΔHfusion​) is the amount of heat required to change a substance from the solid phase to the liquid phase at its melting point. In the heating curve shown, the segment BC represents the phase change from solid to liquid (melting). During this phase change, temperature remains constant while the substance absorbs heat to overcome the forces holding the solid together.

The length of the horizontal segment BC is related to the enthalpy of fusion. A longer BC segment indicates a higher enthalpy of fusion because more heat is required to melt the substance.

10

The process of liquid water freezing at –10 oC and 1 atm pressure is:

Freezing is the process of a liquid turning into a solid, which is exothermic because it releases heat. At –10 oC, which is below the freezing point of water (0 oC), the process is spontaneous because water naturally freezes at temperatures below 0 oC under 1 atm pressure.