AP CHEM WKST KEY: SOLIDS, LIQUIDS, SOLUTIONS REVIEW

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Christine Dixon
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1 ) The particles of solids and liquids are close enough to each other that the interecular forces can exist without any trouble. In gases, the particles are so far apart that the only time IMF s occur is when particles actually collide with each other. However, after the collision the gas particles fly apart and the IMF s no longer exist. 2) a) CH 3OH HB b) HBr DD c) SO 2 DD d) CCl 4 LF e) KCl ionic f) CO 2 LF g) H 2S DD h) NH 3 HB 3) a) H 2O: H 2O is polar with hydrogen bonding while the other two are polar with dipole-dipole. The stronger forces in H 2O will give it the highest boiling point. b) CH 3CH 2CH 3: All 3 of these ecules are nonpolar with London Forces. This ecule is the heaviest and therefore has the strongest forces and highest freezing point. c) CH 3: CH 3CN and CH 3OH are polar and CH 4 is nonpolar. Therefore, CH 4 has the weakest forces and the lowest boiling point. d) HCl: All 3 of these ecules are polar with dipole-dipole forces. HCl is the lightest and therefore has the weakest forces and the lowest freezing point. e) CO: CO is polar with dipole-dipole, the other 2 are nonpolar. Therefore, CO has the strongest forces and the strongest surface tension. f) I 2: All three of these ecules are nonpolar with London Forces. Since I 2 is the heaviest it will have the strongest forces and therefore will have the lowest vapor pressure. g) HOCH 2CH 2OH: HOCH 2CH 2OH is polar with hydrogen bonding whereas the other two ecules are nonpolar with London Forces. Since HOCH 2CH 2OH has the strongest forces, it will have the greatest viscosity. h) NH 3: NH 3 is polar with hydrogen bonding while PH 3 and SbH 3 are both nonpolar with London Forces. Since NH 3 has the strongest forces, it will have the lowest vapor pressure. 4) Since H 2O and glass are both polar, there is an attraction (adhesive force) between the glass and H 2O causing the water to be higher along the outside of its surface while the weight of the water in the middle doesn t allow the cohesive force within the water to pull the middle up even with the outside. Hg is nonpolar and has no adhesive force with the glass. Its inverted meniscus forms because the cohesive forces in it, along with surface tension, cause the middle of the Hg s surface to form a rounded shape. 5) Glass is polar and polyethylene is nonpolar. Since H 2O is polar it will rise farther in the glass where adhesive forces between the two substances exist no such forces exist between the H 2O and polyethylene. The H 2O will rise in the tube until the force of gravity (the weight of H 2O in the tube) is equal to the adhesive force. 6) Liquids form spherical drops because of surface tension and the fact that spheres have the lowest surface area per volume of any of the geometrical 3-dimensional shapes. 7) As long as heat is being added, PE increases at least some. A Since the temperature is rising, the KE of the substance is increasing. B Since temperature is not rising, only the PE of the substance is increasing. C Since the temperature is rising, the KE of the substance is increasing. 8) solid liquid critical temperature critical pressure critical point gas triple point p.

2 9) The solid phase. 0) 82.4 C ) No. The slope of the solid-liquid boundary is not negative nor do the temperatures at the various locations match those of water. 2) a) triple point 4.58 torr, C b) critical point 28 atm, 374C 3) As a liquid freezes it requires some time for the particles to organize in its crystalline form. Since this doesn t even begin to happen until the liquid reaches its freezing point, the temperature will go below the freezing point until the particles get lined up. Once this happens the temperature will rise back up to the freezing point. 4) a) graphite network b) O 2 ecular c) Ba(NO 3) 2 ionic d) Cu metallic 5) a) metallic b) network / ionic c) metallic d) ecular 6) a) NaCl b) S 8 c) Pt e) SF 6 ecular f) Kr Group 8A g) SiO 2 network h) K 2O ionic e) ecular f) metallic g) network i) (NH 4) 2SO 4 ionic j) PCl 3 ecular k) Pt metallic l) H 2Se ecular 7) If injections are not isotonic, the blood cells will be damaged. If injections are hypertonic in regards to the blood, crenation will occur and the blood cells will shrivel up and possibly die. If the injections are hypotonic in regards to the blood, heysis will occur and the cells swell and possibly explode. 8) Only solvent particles will pass through the semipermeable membrane. They will pass in both directions, but the majority of the particles go from the lower concentrated side (lower osmotic pressure) to the more concentrated side (high osmotic pressure). When equilibrium is reached, the same number of solvent particles pass both directions. 9) Solids will generally become more soluble as T increases. Gases will become less soluble as T increases. Liquids generally become more soluble as T increases. 20) Pressure has no effect on the solubility of a solid solute. Pressure has no effect on the solubility of a liquid solute. As pressure increases, the solubility of a gas increases proportionally. 2) CH 3OH, Ba(NO 3) 2, ClF 3, NH 3, CH 3OCH 3, SF 4 22) Nature tends to go in the direction of things with a higher probability. When mixing two substances together, the probability is higher that there will be some type of random mixing of the particles (a disorder of particles) than for the particles to remain separate. 23) There are 3 steps in the dissolving process that requires energy: a) separating solute particles into individual components (expanding the solute) endothermic b) separating solvent particles to make room for the solute (expanding the solvent) endothermic c) the interaction between the solute and solvent to make the solution exothermic or endothermic depending on the polarities of the substances; if polarities are similar: exothermic; if polarities are different: endothermic. Solutions will form if this step is exothermic enough to make ΔH soln negative. Step a above is the reverse of the lattice energy of an ionic substance or the energy to break IMF between ecular compounds. Step b and c will determine the ΔH hyd for the solution. 24) Since the vapor comes from the solvent particles at the surface, the amount of vapor pressure is dependent on the number of solvent particles at the surface. When a nonvolatile solute is added, the solute will replace some of the solute at the surface and p. 2

3 lower the number of solvent particles able to vaporize and thereby lower the vapor pressure for the resulting solution. If a volatile solute is added, the resulting vapor pressure will increase if the solute has a higher vapor pressure than the solvent and will decrease if the solute has a lower vapor pressure than the solvent. 25) A positive deviation from Raoult s law occurs when the P vap is higher than expected mathematically. A negative deviation from Raoult s law occurs when the P vap is lower than expected mathematically. 26) Whether a solution is ideal or have some type of deviation is dependent on the attraction forces between the particles involved in the solution. ideal solution positive deviation negative deviation The solute-solute attraction and the solvent-solvent attraction forces are equal to the attraction forces between solute-solvent. The solute-solute attraction and the solvent-solvent attraction forces are greater then the attraction between solute-solvent. Therefore, the substances will tend to stay in their pure form with weaker forces trying to hold them in the solution. Thus the solution vaporizes more than expected. The solute-solute attraction and the solvent-solvent attraction forces are weaker than the forces between the solute-solvent. The stronger forces between solutesolvent will hold the substances in the solution more readily than the substances are able to vaporize. Thus, the solution doesn t vaporize as much as expected. ΔH 0 ΔH > 0 ΔH < 0 27) nonvolatile, nondissociating solute the P soln will be lower than the P vap of the pure solvent nonvolatile, dissociating solute the P soln will be lower than the P vap of the pure solvent volatile solute with P vap lower than the P vap of solvent the P soln will be lower than the P vap of the pure solvent volatile solute with P vap higher than the P vap of solvent the P soln will be higher than the P vap of the pure solvent 28) The freezing point and boiling point of a liquid is dependent on its P vap. By adding a solute to water, the P soln will be lower than the P vap of pure water. Therefore, the freezing point of the solution will be less than that of pure water because less heat is needed to reach the equilibrium point between the P vap of a solid and a liquid. However, the boiling point will increase because now more heat is needed to reach the equilibrium point between the P vap of a liquid and a gas. 29) Since NaCl is ionic it will dissociate into its ions when dissolved and have a i2 (if we don t worry about the van t Hoff factor).. So, a.0 m NaCl solution will have 2.0 m of solute particles. Sugar is ecular and doesn t dissociate when dissolved. Therefore it has i. So, the sugar solution will have.0 m of solute particles. Therefore, the NaCl solution will cause ΔT b, since this is a colligative property, to be greater than the sugar solution. 30) When the NaCl dissolves it dissociates into its ions. We would expect twice as much solute particles than the number of NaCl units dissolved. However, the ions are in constant motion in the solution and will come in contact with oppositely charged ions from time to time. This contact will cause ion pairs to temporarily form. Therefore, at any given point in time, the total number of solute particles in solution will be less than expected, thus the temperature changed is not as great as expected. 3).0 M C 6H 2O 6.0 M solute particles.0 M NaCl 2.0 M solute particles.0 M MgCl M solute particles.0 M (NH 4) 3PO M solute particles.0 M C 6H 2O 6 >.0 M NaCl >.0 M MgCl 2 >.0 M (NH 4) 3PO 4 32) Reverse osmosis occurs when a pressure greater than the osmotic pressure of the solution is applied to the solution. This will force the solvent particles back across the semipermeable membrane into the pure solvent. An example is the desalination of ocean water to produce drinkable water. 33) The blood passes through a cellophane tube which acts as a semipermeable membrane. The tubing is placed into a dialyzing solution (a washing solution ) which contains the same concentration of ions and small ecules as blood but has none of the waste products normally removed by the kidneys. The waste products in the blood move into the washing solution and thus clean the blood. 34) Substitutional alloys are when a solid solute replaces a particles of the solid solvent in the lattice structure. Interstitial alloys occur when a solid solute particle fits into the interstices (the empty spaces) of the solid solvent s structure. 35) a) colloid b) suspension c) solution d) colloid p. 3

4 36) Heating the mixture or adding an electrolyte. Heat will cause the particles to move around with more energy and can overcome the repulsion forces caused by the ion ring around the dispersed particles. Adding an electrolyte will neutralize the ion ring so the dispersed particles can coagulate. 37) 42.50% KOH / d soln.225 g/ml -assume 00 g soln g KOH KOH g H 2O 3.9 H 2O V soln m soln d soln 00 g.225 g ml 8.63 ml soln M L soln L M KOH X KOH KOH m kg solvent 3.7 m KOH kg N nm ()(9.278 M) N KOH 38) 3.50 m H 2CO 3 / d soln.05 g/ml -assume kg H 2O 3.50 H 2CO 3 27 g H 2CO H 2O 000 g H 2O g soln V soln m soln d soln 27 g 0 ml soln.05 g ml M L soln L 3.8 M H 2CO 3 X H2 CO 3 H 2CO mass solute 27 g % H 2 CO 3 ( ) 00 ( mass soln 27 g ) % H 2CO 3 N nm (2)(3.8 M) 6.36 N H 2CO 3 39) M K 2SO 4 / d soln.008 g/ml / SO 4 2 SO assume L soln K 2SO 4 26 g K 2SO H 2O 882 g H 2O 49.6 m kg solvent kg m K 2SO 4 m soln d solnv soln (.008 g/ml)(000 ml) 008 g soln X K2 SO 4 K 2SO mass solute 26 g % K 2 SO 4 ( ) 00 ( mass soln 008 g ) % K 2SO 4 N nm (2)(0.725 M).45 N K 2SO 4 40) 35.0 g KNO 3 / 445 ml soln / T 37.0 C g solute Π [ (MM solute)(l soln) ] RTi [ 35.0 g L atm (0. g ( ) (30.2 K)(2) 2.2 atm )(.445 L)] K 4) g C 6H 2O 6 / g CS 2 / FP CS C; K f 3.83 C/m / BP CS C; K b 2.34 C/m g m (80.8 g.0 m )( kg) p. 4

5 ΔT f K fmi (3.83 C/m)(.0 m)() 4.25 C FP -.5 C 4.25 C -5.8 C ΔT b K bmi (2.34 C/m)(.0 m)() 2.60 C BP 46.2 C C 48.8 C 42) P C 6H torr / P C 7H torr / P CH 3OH torr 50.0 g C 6H C 6H g C 7H C 7H g CH 3OH.34 CH 3OH 2.37 total P soln P C6 H 6 Χ C6 H 6 + P C7 H 8 Χ C7 H 8 + P CH3 OHΧ CH3 OH P soln (93.4 torr) ( ) + (26.9 torr) (0.39 ) + (224.5 torr) ( ) P soln 25.2 torr torr + 27 torr 57 torr 43) P vap T atm / P vap T2 atm / T C / ΔH vap kj/ P vap T P vap T2 ) H vap R ( T atm atm ) J T ) J K ( x K ) (3482 K) ( x K- ) 6.40 x 0 5 K x K K - x x K C 44) T C / T 2 00 C / P vap T2 atm / ΔH vap 40.7 kj/ P vap T P vap T2 ) H vap R ( T 2 - x atm ) J x atm ) -.92 x 0.47 atm J K T ) ( K K ) p. 5

AP CHEM WKST KEY: SOLIDS, LIQUIDS & GASES REVIEW

AP CHEM WKST KEY: SOLIDS, LIQUIDS & GASES REVIEW 1) a) NaCl ionic b) C6H12O6 HB c) PH3 LF d) H2O2 HB e) CH3CH2OH HB f) HI DD g) Kr LF h) diamond covalent 2) a) H2O Want the strongest IMF for highest BP.