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# Chap 14 practice problems answers - C 5 H 10 What is the...

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EXAM PRACTICE QUESTIONS Chapter 14 1. How many grams of sucrose (C 12 H 22 O 11 ) must be dissolved in 750. mL of water to prepare a 0.250 molal solution? 64.1 g 2. If the mole fraction of CH 3 OH in a solution with only water is 0.0250, what is the molality of the CH 3 OH? 1.42 m 3. In a home ice cream freezer, we lower the freezing point of the water bath surrounding the ice cream by dissolving NaCl in water to make a brine solution. A 15.0% brine solution is observed to freeze at -10.888 ° C. What is the van't Hoff factor, i , for this solution? 1.94 4. Indicate (by writing yes or no in the blanks below) which of the following pairs will form a solution when mixed. KI in CCl 4 LiF in water no CH 3 OH in water yes CH 3 OH in C 6 H 6 yes vegetable oil in mineral oil no yes 5. What is the vapor pressure of an aqueous solution containing 10.0% ( w/w ) ethylene glycol (62.0 g/mol) at 25 ° C. P H2O = 24.3 torr at 25 ° C. 23.5 torr 6. At 25 ° C a solution consists of 0.450 mole pentane, C 5 H 14 , and 0.250 mol of cyclopentane,

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Unformatted text preview: C 5 H 10 . What is the lowering of the vapor pressure of pentane in this solution? The vapor pressure of the pure liquids at 25 ° C are 451 torr for pentane and 321 torr for cyclopentane. 161 torr 7. When 1.150 grams of an unknown electrolyte dissolves in 10.0 grams of water, the solution freezes at -2.16 ° C. What is the molecular weight of the unknown compound? K f for water is 1.86 ° C/m. 99.1 g/mol 8. The osmotic pressure of 1.00 L of an aqueous solution of a nonelectrolyte solute is 1.17 atm at 0.00 ° C. The density of the solution is 1.09 g/mL. The chemical formula of the nonelectrolyte is C 3 H 6 O 3 . What is the % m/m of the nonelectrolyte in the solution? 0.431% m/m 9. Which solution would produce the lowest freezing point? 0.075 m sodium sulfate 10. A 0.0490 m aqueous NaBr solution freezes at -0.173 ° C. What is the apparent percent dissociation if NaBr in this solution? K f = 1.86 ° C/m for water NaBr → Na + + Br-89.8%...
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