CHM1 11 Molality and Mole Fractions Collection

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Molality and Mole Fractions

From ChemPRIME

Molality

The molality (m) of a solution is used to represent the amount of moles of solute per kilogram of the solvent.

$m = \frac{Moles of Solute}{Kilograms of Solvent}$ $m = ("Moles of Solute")/("Kilograms of Solvent")$

The molarity and molality equations differ only from their denominators. However, this is a huge difference. As you may remember, volume varies with different temperatures. At higher temperatures, volumes of liquids increase, while at lower temperatures, volumes of liquids decrease. Therefore, molarities of solutions also vary at different temperatures. This creates an advantage for using molality over molarity. Using molalities rather than molarities for lab experiments would best keep the results within a closer range. Because volume is not a part of its equation, it makes molality independent of temperature.

EXAMPLE 4 0.88g NaCl is dissolved in 2.0L water. Find its molality.

Mass and Mole Fractions

When speaking of solubility or miscibility, or when doing quantitative experiments involving solutions, it is necessary to know the exact composition of a solution. This is invariably given in terms of a ratio telling us how much solute is dissolved in a unit quantity of solvent or solution. The ratio can be a ratio of masses, of amounts of substances, or of volumes, or it can be some combination of these. For example, concentration was defined before as the amount of solute per unit volume of solution:

$C_{solute} = \frac{n_{solute}}{V_{solution}}$ $C_"solute" = (n_"solute")/(V_"solution")$

The two simplest measures of the composition are the mass fraction w, which is the ratio of the mass of solute to the total mass of solution, and the mole fraction x, which is the ratio of the amount of solute to the total amount of substance in the solution. If we indicate the solute by A and the solvent by B, the mass fraction and the mole fraction are defined by

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EXAMPLE 5 A solution is prepared by dissolving 18.65 g naphthalene, C₁₀H₈ in 89.32 g benzene, C₆H₆. Find (a) the mass fraction and (b) the mole fraction of the naphthalene.

The mass fraction is useful because it does not require that we know the exact chemical nature of both solute and solvent. Thus if we dissolve 10 g crude oil in 10 g gasoline, we can calculate w _{crude oil}= 0.5 even though the solute and solvent are both mixtures of alkanes and have no definite molar mass. By contrast, the mole fraction is useful when we want to know the nature of the solution on the microscopic level. In the above example, for instance, we know that for every 100 mol of solution, 11.28 mol is naphthalene. On the molecular level this means that out of each 100 molecules in the solution, 11.28 will, on the average, be naphthalene molecules.

Equations

Mole fraction
$x_{a} = \frac{n_{a}}{n_{a} + n_{b}}$ $x_a = n_a/(n_a+n_b)$
Dasha Use Equation
Mass Fraction
$w_{a} = \frac{m_{a}}{m_{a} + m_{b}}$ $w_a = m_a/(m_a+m_b)$
Dasha Use Equation
Molality
$Molality (m) = \frac{n_{solute}}{m_{solvent}}$ $"Molality" (m) = n_"solute" / m_"solvent"$
DavidC Use Equation
Molarity
$M = \frac{(m) moles solute}{(k) volume solution}$ $M = "(m) moles solute" / "(k) volume solution"$
ekskekel Use Equation

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CHM1 11 Molality and Mole Fractions Collection

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