Chem 1C Test 2
Ch. 17 56 Missed The two beakers in the sealed container illustrated below contain pure water and an aqueous solution of a volatile solute. (Image) If the solute is less volatile than water, explain what will happen to the volumes in the two containers as time passes.
Because the solute is volatile, both the water and solute will transfer back and forth between the two beakers. The volume in each beaker will become constant when the concentrations of solute in the beakers are equal to each other. Because the solute is less volatile than water, one would expect there to be a larger net transfer of water molecules into the right beaker than the net transfer of solute molecules into the left beaker. This results in a larger solution volume in the right beaker when equilibrium is reached, i.e., when the solute concentration is identical in each beaker.
Ch. 17 55 Missed When pure methanol is mixed with water, the solution gets warmer to the touch. would you expect this solution to be ideal? Explain
No, the solution is not ideal. For an ideal solution, the strengths of intermolecular forces in the solution are the same as in pure solute and pure solvent. This results in ΔHsoln = 0 for an ideal solution. ΔHsoln for methanol-water is not zero. Because ΔHsoln < 0, this solution exhibits negative deviation from Raoult's law.
Ch. 17 79 Missed A solution of sodium chloride in water has a vapor pressure of 19.6 torr at 25C. What is the mole fraction of NaCl in this solution? What would be the vapor pressure of this solution at 45C? The vapor pressure of pure water is 23.8 torr at 25C and 71.9 torr at 45C.
P= χP°; 19.6 torr = χH2O (23.8 torr), χH2O = 0.824; χsolute = 1.000 - 0.824 = 0.176 0.176 is the mole fraction of all the solute particles present. Because NaCl dissolves to produce two ions in solution (Na+ and Cl−), 0.176 is the mole fraction of Na+ and Cl− ions present. The mole fraction of NaCl is 1/2 (0.176) = 0.0880 = χ NaCl . At 45°C, Psoln = 0.824(71.9 torr) = 59.2 torr.
Ch. 17 52 Missed Match the vapor pressure diagrams with the solute-solvent combinations and explain your answers. (Image needed)
The first diagram shows positive deviation from Raoult's law. This occurs when the solute- solvent interactions are weaker than the interactions in pure solvent and pure solute. The second diagram illustrates negative deviation from Raoult's law. This occurs when the solute- solvent interactions are stronger than the interactions in pure solvent and pure solute. The third diagram illustrates an ideal solution with no deviation from Raoult's law. This occurs when the solute-solvent interactions are about equal to the pure solvent and pure solute interactions. These two molecules are named acetone (CH3COCH3) and water. As discussed in section 17.4 on nonideal solutions, acetone-water solutions exhibit negative deviations from Raoult's law. Acetone and water have the ability to hydrogen bond with each other, which gives the solution stronger intermolecular forces as compared to the pure states of both solute and solvent. In the pure state, acetone cannot H−bond with itself. So the middle diagram illustrating negative deviations from Raoult's law is the correct choice for acetone- water solutions. These two molecules are named ethanol (CH3CH2OH) and water. Ethanol-water solutions show positive deviations from Raoult's law. Both substances can hydrogen bond in the pure state, and they can continue this in solution. However, the solute-solvent interactions are somewhat weaker for ethanol-water solutions due to the significant nonpolar part of ethanol (CH3−CH2 is the nonpolar part of ethanol). This nonpolar part of ethanol slightly weakens the intermolecular forces in solution. So the first diagram illustrating positive deviations from Raoult's law is the correct choice for ethanol-water solutions. These two molecules are named heptane (C7H16) and hexane (C6H14). Heptane and hexane are very similar nonpolar substances; both are composed entirely of nonpolar C−C bonds and relatively nonpolar C−H bonds, and both have a similar size and shape. Solutions of heptane and hexane should be ideal. So the third diagram illustrating no deviation from Raoult's law is the correct choice for heptane-hexane solutions. These two molecules are named heptane (C7H16) and water. The interactions between the nonpolar heptane molecules and the polar water molecules will certainly be weaker in solution as compared to the pure solvent and pure solute interactions. This results in positive deviations from Raoult's law (the first diagram).
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