Explanation: Concentration tells us how much of one component (in a mixture) we have relative the mixture (or another component in the mixture). That is, for a mixture of sugar and water, a concentration can specify how much water we have relative to the solution (water and sugar), or how much water we have relative to the other component (sugar). These concentrations can be expressed using units of amount (mols, molecules, ...), mass (kg, lb, ...), and volume (m3, L, ...). The following are some ways to express concentration, where the numerator signifies the solute and the denominator signifies the solution. The last molar concentration listed, g-mol/L is the Molarity of the solute in the mixture. Thus, the definition of molarity follows: Molarity of component A, Mass and mole fractions will be used frequently in material and energy balance problems. The notation used in the course and thus the notation used here differs from the notation introduced in chapter 3 (here, x designates a liquid or solid mass or mole fraction, and y designates mass or mole fraction for a gas). Here are the definitions for mass and mole fractions: Equation 1: Equation 2: Equation 3: Equation 4: xA, yA - mass or mole fractions of component A.
"x" is used for liquids and solids, thus the first two equations are the mass and mole fractions for liquids and solids. Accordingly, the last two equations refer to gases. As you can see, "xA" can denote several different things. For example, xH2O can refer to either the mass or mole fraction of water or ice in a mixture. It will be upto the problem solver to keep track of how "xA" is defined in working a problem. Related to the mass and mole fractions are the units of parts per million (ppm) and the parts per billion (ppb). Here, we simply multiply the mass or mole fraction by 106 for ppm, and by 109 for ppb. parts per million parts per billion note: The mole and mass fractions must always be unitless, so the mole/mass of the component (numerator) and the total mole/mass (denominator) must both have the same units. Example 1 A mixture of gases has the following composition by mass: O2 16% CO 4.0% CO2 17% N2 63%And, for mass fractions, we would move the decimal place over two places, so, for example, the mixture is 16% O2, or we could say, the mole fraction of O2 in the mixture is .16. Determine the molar composition of this mixture. (hint: you first need to find the total moles of the mixture) Example 2 Calculate the average molecular weight of air. Air is 79% N2 and 21% O2, and use the following formula to calculate the average molecular weight: Example 3 A material balance problem (chapter 4) gives us the following quantities of compounds going into a condenser. Determine the mole fractions of each stream.PLACE "CONCEN BMP" HERE |
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