1.1 What is Chemistry?

http://www.ck12.org Chapter 16. Solutions

between molarity and molality. Molality uses mass instead of volume, and we are looking at the amount ofsolvent
instead of the total amount ofsolution. It can be calculated as follows:

molality =moles solutekg solvent

Molality is primarily useful for calculating certain physical properties of solutions, such as freezing point, boiling
point, and vapor pressure. These will be discussed in the next lesson.

Example 16.10

What would be the molality of an sugar solution in which 4.00 g C 6 H 12 O 6 was dissolved in 1.00 L of water? Water
has a density of 1.00 g/mL.

Answer:

molality=moles solutekg solvent

We need to find the amount of solute (sugar) in moles and the mass of the solvent in kilograms.

moles solute= 4 .00 g C 6 H 12 O 6 × 342 1 mol C.30 g C^6 H 6 H^1212 O^6 O 6 = 0 .0117 mol C 6 H 12 O 6

kg solvent= 1 .00 L H 2 O×1000 mL H1 L H 2 O^2 O×^1 1 mL H.00 g H 22 OO×1000 g1 kg = 1 .00 kg H 2 O

Substituting these values into our molality expression, we get the following:

molality=^0 .0117 mol NaCl 1 .00 kg H 2 O = 0 .0117 m

Weight Percent

Another way to express the concentration of a solution is by its weight percent. This is commonly used to describe
stock solutions of things like acids and bases.Weight percentcan be calculated by dividing the mass of a solute by
the mass of the entire solution:

Weight Percent=grams of solutiongrams of solute ×100%

A solution in which 9.3% of the mass is due to NaCl would be referred to as a 9.3% (w/w) NaCl solution. The
presence of "(w/w)" indicates that the ratio is between the weight (mass) of the solute and the weight of the total
solution. Other ratios in which one or both of the components are measured in terms of volume instead of mass are
also common; these percentages are labeled as either weight/volume (w/v) or volume/volume (v/v).

Example 16.11

What is the weight percent of a solution that has 9.01 g of NaCl dissolved in 1000. g of water?

Answer:

Grams of solution = grams of NaCl + grams of water = 9.01 g NaCl + 1000 g H 2 O = 1009 g solution

Weight Percent=1009 g solution^9 .01 g NaCl ×100%= 0 .893%

This mixture could be described as a 0.893% (w/w) solution of NaCl in water.

Parts Per Million

We can also express concentrations by dividing the number of particles of a certain solute by the total number of
particles in a solution. These types of values are commonly used to describe small amounts of a substance in a