Victory for the ACT Student Text 15e

L ESSON 2 | R ESEARCH S UMMARY • 315

PASSAGE V Freezing point depression occurs when a non-volatile solute is added to a pure solvent, resulting in a solution with a lower freezing point than that of the pure solvent. The freezing point depression, T , T is roughly proportional to the concentration of the solute in the solution (as measured by molality): T K m i f # # T (Equation 1) K f is the freezing point depression constant for the solvent, m is the molality of the solution (moles of solute per kilogram of solvent), and i is the van’t Hoff factor, which is equal to the number of particles produced by dissolving one unit of solute. A chemistry student uses the phenomenon of freezing point depression to determine the ‘Žƒ” ƒ•• ƒ† ‹†‡–‹ϐ‹…ƒ–‹‘ ‘ˆ ƒ —‘™ •—„•–ƒ…‡ ˆ”‘ ƒ„Ž‡ ͳǤ Table 1 Chemical Molar Mass (grams/mole) Sodium iodide (NaI) 150 Calcium sulfate (CaSO 4 ) 136 Sodium nitrate (NaNO 3 ) 85 Sodium chloride (NaCl) 58 ‘†‹— ϐŽ—‘”‹†‡ ȋ ƒ Ȍ 42 Part 1 –‡•– –—„‡ ϐ‹ŽŽ‡† ™‹–Š †‹•–‹ŽŽ‡† ™ƒ–‡” ‹• ’Žƒ…‡† ‹ ƒ •ƒŽ–Ȁ‹…‡ ™ƒ–‡” „ƒ–Š ƒ– „‡Ž‘™ 10 q C and constantly stirred until crystals begin to form, at which point the distilled water temperature is recorded. The procedure is repeated for a sugar-water solution of molality 1.0 m (1.0 mol sugar/1 kg water) and van’t Hoff factor of 1. The student uses the freezing point depression for the sugar- water solution to calculate the freezing point depression constant for water (Table 2). Table 2 1. Freezing point of distilled water –0.04 q C 2. Freezing point of 1.0 m sugar-water solution –1.91 q C ' T for sugar-water solution –1.87 q C 4. K f value for distilled water –1.87 q C/(mol/kg) Part 2 A beaker is weighed before and after the addition of approximately 100 mL of distilled water. The beaker is weighed again after the addition of a sample of an unknown substance from Table 1 known to have a van’t Hoff factor of 2. The student calculates the solution concentration (grams of solute per kilogram of water). The freezing point depression of the unknown solution is determined as in Part 1, and the solution molality is calculated. Finally, the student calculates the molar mass of the unknown solute using Equation 2. molar mass (grams/mole) molality (moles/kilogram of water) concentration (grams/kilogram of water) (Equation 2)

Throwback to Science If you are confused about how the units work out in Equation 2, you can set up the units in a dimensional analysis equation. Just take the unit fractions of the numerator and denominator, making •—”‡ –‘ ϐŽ‹’ –Š‡ ˆ”ƒ…–‹‘ of the denominator since it is on bottom, and multiply them together. Then cancel out the matching units, which will leave you with the units of –Š‡ ϐ‹ƒŽ ƒ•™‡”Ǥ ‘” example: kg of water g mol kg of water

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