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Solids and Liquids Study Guide (page 2)

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Updated on Sep 25, 2011

Example:

Which compound of the following pairs would you expect to have the higher melting point?

Compounds Higher melting point (also higher boiling point)
Ne and Kr Kr; only van der Waals forces are present in the pair and Kr is the largest atom.
NH3 and NCl3 NH3; both are polar molecules, but only ammonia (NH3) has hydrogen bonding.
CO2 and NO2 NO2; carbon dioxide is linear and nonpolar (only van der Waals forces), and nitrogen dioxide is bent and polar.

Colloids

Colloids are stable mixtures in which particles of rather large sizes (ranging from 1 nm to μm) are dispersed throughout another substance. Aerosol (liquid droplets or solid particles dispersed in a gas) such as fog can scatter a beam of light (Tyndall effect).

Colligative Properties

Colligative properties are solution properties that vary in proportion to the solute concentration and depend only on the number of solute particles. This section covers a few solubility laws based on colligative properties.

Henry's Law (a gas-liquid solution) states that the solubility of an ideal gas (C) in mol per liter is directly proportional to the partial pressure (p) of the gas relative to a known constant (k) for the solvent and gas at a given temperature.

C = kp

Raoult's Law (a solid-liquid solution) says that the vapor pressure of an ideal solution (ptotal) is directly proportional to the partial vapor pressure (pA) of the pure solvent times the mole fraction (XA = moles of solute per moles of solute and solvent) of the solute.

ptotal = XApA

Raoult's Law (a liquid-liquid solution) states that the vapor pressure of an ideal solution of two liquids (ptotal) is directly proportional to the vapor pressures of the pure liquids, the mole fractions of the liquids (XA and XB), and the partial vapor pressure (pA and pB) of the liquids above the solution.

Example:

Use Henry's Law to calculate the solubility of oxygen (~21% of the atmosphere) in water at STP (273.15 K and 1.00 atm). The Henry's Law constant for O2(g) at 273.15 K is 2.5 * 10–3 M.

Solution:

  • Calculate p: p = X * (1.00 atm) = 0.21 atm O2
  • Calculate concentration (C): C = kp =

Boiling Point Elevation

Solutions containing nonelectrolyte nonvolatile solutes have higher boiling points than the pure solvent. The boiling point elevation (Tb) is directly proportional to the solvent's boiling point elevation constant (Kb) times the molality (m) of the solute in moles per kg of solvent:

Tb = Kbm

Freezing Point Depression

Solutions have lower freezing points than the pure solvent. The freezing point elevation (Tf) is directly proportional to the solvent's freezing point depression constant (Kf) times the molality (m) of the nonelectrolyte solute in moles per kg of solvent:

Tf = Kfm

Table 14.1 Common Boiling Point Elevation (Kb) and Freezing Point Depression (Kf) Constants

 

Example 1:

Calculate the freezing point of a sugar solution of 0.34 mol of C6H12O6 in 275 g of water.

Solution 1:

1.
2. Find Tf : Tf = Kf m = 1.86 * 1.2363 = 2.3° C
3. The normal freezing point of water is 0° C, so 0° C – 2.3° C = –2.3° C.
4. A practical example of freezing point depression is the use of salt on icy roads. Salt does not "melt" the ice; it just lowers the freezing point.

 

Example 2:

Calculate the boiling point of a solution of 1.25 mol of sucrose (table sugar) dissolved in 1,550 g of water.

Solution 2:

1.
2. Find Tb: Tb = Kbm = 0.512 * 0.80645 = 0.413° C
3. The normal boiling point of water is 100° C, so 100° C + 0.413° C = 100.413° C.

 

Practice problems for these concepts can be found at -  Solids and Liquids Practice Questions

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