You have made up a solution of known molarity but now realize that you need to know the molality instead. Find the molality of sucrose, C₁₂H₂₂O₁₁, in 1.06 m C₁₂H₂₂O₁₁(aq), which is known to have density 1.140 g · mL^–1.

ANTICIPATE The mass of 1 L of aqueous solution is close to 1 kg, so the numerical value of the molality can be expected to be close to that of the molarity, but with different units, of course.

PLAN Use procedure 3 in Toolbox 5E.1, taking note of the discussion of units.

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Toolbox 5E.1 HOW TO USE THE MOLALITY CONCEPTUAL BASIS The molality of a solute is its concentration in moles per kilogram of solvent. Its value is independent of the temperature and is directly proportional to the relative numbers of solute and solvent molecules in the solution. To convert molarity to molal- ity, note that the former is defined in terms of the volume of the solution; thus, you need the density of the solution to convert that overall volume to the mass of solvent present. PROCEDURE When working with molality, keep its definition (b = nsolute/msolvent) in mind. The expressions derived here are sum- marized in TABLE 5E.1. 1. Calculating the mass of solute in a given mass of solvent from the molality Step 1 Calculate the amount of solute molecules, nsolute present in a given mass of solvent, msolvent by converting the mass of solvent from grams to kilograms, if necessary, and rearranging the equation defining molality (Eq. 3) into nsolute = bx msolvent 2. Calculating the molality from a mole fraction Step 1 Consider a solution composed of a total amount of molecules, n. If the mole fraction of the solute is xsolute, the amount of solute molecules present is 1solute = Xsolute X n Step 2 If there is only one solute, the mole fraction of solvent molecules is 1 - Xsolute. The amount of solvent molecules is then nsolvent = (1 - Xsolute)n. Convert this amount into mass by using the molar mass of the solvent, Msolvent m solvent = nsolvent Msolvent = {(1 Step 3 From Eq.3, b = - Xsolute) n} Msolvent Xsolute Xsoluten (1-xsolute)n M solvent (1 - Xsolute) Msolvent A note on units: The mole fraction solute is a dimensionless num- ber. Therefore b has the same units as 1/Msolvent Express Msolvent in kilograms per mole (kg mol¹) by div its usual value (in grams per mole) by kg/g = 10³ (so a molar mass of 55 g-mol™¹ becomes 0.055 kg-mol¹). Then the molality b is obtained in moles per kilogram (mol kg ¹). This expression can be rearranged to express the mole fraction