1. Hints and Example. Use the definitions of molality, weight percent, and mole fraction:

molality = moles solute / kg solvent

wt. % = mass solute / 100 g solution

(remember that mass of solution = mass of solvent + mass of solute)

mole fraction of X = moles of X / total moles in the mixture

Answer: The solution contains 1 mole of NaCl in exactly 1000 g solvent. or 58.44 g NaCl in 1000 g water.

Therefore the total mass = 58.44 g NaCl + 1000 g H₂O = 1058.44 g

Wt. % NaCl = 100 * 58.44 g/1058.44 g = 5.52%

Mole fraction = moles NaCl / (total moles, H₂O + NaCl)

moles H2O = 1000 g * 1 mol/18.02g = 55.5 moles H₂O

mole fraction of NaCl = 1.00 mol NaCl /( 1.00 mol NaCl + 55.5 moles H₂O)

= 0.0177

(a) Exothermic processes release energy to the surroundings, endothermic processes take energy from the surroundings. Review Chapter 6: relationship of sign of energy change to endothermic/exothermic reactions, and page 655.

(b) See page 655, Temperature Effects on Solubility and LeChatelier's principle

(c) See example on page 651, where the standard enthalpy of formation of aqueous NaCl is calculated from the standard enthalpy of formation of solid NaCl and the enthalpy of solution of NaCl.
 
 

This is a Henry's law calculation; Henry's law constants are found on p. 653, and an example is shown on p. 654.
 

Get molality of solution: moles of solute / mass of solvent in kg. Then use the equation for boiling point elevation, with the appropriate value of K_b for the solvent (p. 661). See worked example on p. 662.
 

Calculate the total molality of the ions in each case, using the van't Hoff factor (# of ions in one formula unit of the compound). Remember that covalent molecules such as CH₃OH do not dissociate into ions, so their van't Hoff factor is 1.

Then recall that the freezing/boiling point is directly proportional to the total molality of particles present. Solutions with the highest molality have the greatest boiling point elevations and freezing point depressions.

1 ppm = 1 g solute in 1,000,000 g solution

= 1 mg solute in 1000 g solution

= 1 mg solute in 1 liter solution if the density of the solution = 1.00 g/mL

So, a solution that is 312 ppm in some solute X contains 312 mg of X per liter.
 
 

Use the equation for osmotic pressure.

All the variables are known except the concentration c (molarity). Find c, then use the given volume to determine the number of moles present in the solution:

molarity x volume = moles

Finally get the molar mass from the moles and the mass of the solute.
 

1 atm = 760 mm Hg. In terms of water (the tree sap), this corresponds to

760 mm Hg x 13.1 mm H₂O = 9.96 x 10³ mm H₂O

Now you can determine what the osmotic pressure (in atm) has to be; it's the height of the tree divided by 1 atm pressure expressed as mm H₂O.
 

Freezing point is determined by molality, vapor pressure by mole fraction (Raoult's Law). See page 658 for a worked example of Raoult's law.
 

How many grams of the salt can be dissolved in the given amount of water at the lower temperature?

How many grams of the salt can be dissolved in the given amount of water at the higher temperature?

The difference in these two numbers is how much salt will precipitate when the solution is cooled.
 
 

This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem181/CHEM_1P81_Assign_5.html
Last revised: February 14 2001 by M. F. Richardson
© Brock University, 2001