基礎(chǔ)化學(xué)第二章后習(xí)題答案.doc

資料收集于網(wǎng)絡(luò) 如有侵權(quán)請聯(lián)系網(wǎng)站 刪除 謝謝 第二章后習(xí)題答案 習(xí)題1. 水在20時(shí)的飽和蒸氣壓為2.34 kPa 。若于100g水中溶有10.0 g蔗糖(Mr= 342)，求此溶液的蒸氣壓。解 根據(jù) , 2. 甲溶液由1.68 g蔗糖(Mr=342)和20.00 g水組成，乙溶液由2.45 g （Mr= 690）的某非電解質(zhì)和20.00 g水組成。在相同溫度下，哪份溶液的蒸氣壓高？將兩份溶液放入同一個(gè)恒溫密閉的鐘罩里，時(shí)間足夠長，兩份溶液濃度會不會發(fā)生變化，為什么？當(dāng)達(dá)到系統(tǒng)蒸氣壓平衡時(shí)，轉(zhuǎn)移的水的質(zhì)量是多少？解 (1) 溶液乙的蒸氣壓下降小，故蒸氣壓高。（2）乙溶液濃度變濃, 甲溶液濃度變稀。因?yàn)闈舛炔煌娜芤褐糜谕幻荛]容器中，由于不同，P不同， 蒸發(fā)與凝聚速度不同。乙溶液蒸氣壓高，溶劑蒸發(fā)速度大于甲溶液蒸發(fā)速度，所以溶液乙中溶劑可以轉(zhuǎn)移到甲溶液。（3）設(shè)由乙溶液轉(zhuǎn)移到甲溶液的水為x(g), 當(dāng)兩者蒸氣壓相等時(shí)，則x = 3.22g3.將2.80 g難揮發(fā)性物質(zhì)溶于100 g水中，該溶液在101.3 kPa下，沸點(diǎn)為100.51 。求該溶質(zhì)的相對分子質(zhì)量及此溶液的凝固點(diǎn)。(Kb = 0.512 Kkgmol-1，Kf = 1.86Kkgmol-1)解 該溶液的凝固點(diǎn)為-1.854.煙草有害成分尼古丁的實(shí)驗(yàn)式是C5H7N，今將538 mg尼古丁溶于10.0 g水，所得溶液在101.3 kPa下的沸點(diǎn)是100.17 。求尼古丁的分子式。解 尼古丁的分子式為： 5. 溶解3.24 g硫于40.0 g苯中，苯的凝固點(diǎn)降低1.62。求此溶液中硫分子是由幾個(gè)硫原子組成的？(Kf = 5.10 Kkgmol-1 )解 此溶液中硫原子是由8個(gè)硫原子組成。6. 試比較下列溶液的凝固點(diǎn)的高低：(苯的凝固點(diǎn)為5.5 ，Kf = 5.12 Kkgmol-1，水的Kf = 1.86 Kkgmol-1) 0.1 molL-1蔗糖的水溶液； 0.1 molL-1乙二醇的水溶液； 0.1 molL-1乙二醇的苯溶液； 0.1 molL-1氯化鈉水溶液。解 對于非電解質(zhì)溶液，電解質(zhì)溶液，故相同濃度溶液的凝固點(diǎn)的大小順序是： 7. 試排出在相同溫度下，下列溶液滲透壓由大到小的順序： c(C6H12O6)= 0.2 molL-1； ； ； c(NaCl)= 0.2 molL-1。解 根據(jù)非電解質(zhì)溶液, 電解質(zhì)溶液,滲透壓大小順序是： 8. 今有一氯化鈉溶液，測得凝固點(diǎn)為 -0.26 ，下列說法哪個(gè)正確，為什么？ 此溶液的滲透濃度為140 mmolL-1； 此溶液的滲透濃度為280 mmolL-1； 此溶液的滲透濃度為70 mmolL-1； 此溶液的滲透濃度為7 153 mmolL-1 。解 由于 NaCl 在水溶液中可以電離出2倍質(zhì)點(diǎn)數(shù)目，該溶液的滲透濃度可認(rèn)為： 所以（1）正確，氯化鈉溶液的滲透濃度應(yīng)為140 mmolL-19. 100 mL水溶液中含有2.00 g 白蛋白，25 時(shí)此溶液的滲透壓力為0.717 kPa 求白蛋白的相對分子質(zhì)量。解 10. 測得淚水的凝固點(diǎn)為 -0.52 ，求淚水的滲透濃度及 37 時(shí)的滲透壓力。解 淚水的滲透濃度為。11.今有兩種溶液，一為1.50 g 尿素(Mr = 60.05)溶于200 g 水中，另一為42.8 g 某非電解質(zhì)溶于1000 g 水中，這兩種溶液在同一溫度下結(jié)冰，試求該非電解質(zhì)的相對分子質(zhì)量。解 若兩溶液在同一溫度下結(jié)冰，則 ，有 12. 在0.100kg的水中溶有0.020 mol NaCl, 0.010 mol Na2SO4和0.040 mol MgCl2。假如它們在溶液中完全電離，計(jì)算該溶液的沸點(diǎn)升高值。解 他們在溶液中完全電離，溶液中總質(zhì)點(diǎn)數(shù)目為： Exercises1. What are the normal freezing points and boiling points of the following solution?(a)21.0g NaCl in 135mLof water.(b) 15.4g of urea in 66.7 mL of water.Solution: （a） (b) at ease 舒適；快活；自由自在2. If 4.00g of a certain nonelectrolyte is dissolved in 55.0g of benzene, the resulting solution freezes at 2.36. Calculate the molecular weight of the nonelectrolyte. Solution: 3. The average osmotic pressure of seawater is about 30.0 atm at 25. Calculate the concentration (molarity) of an aqueous solution of urea (NH2CONH2)that is isotonic with seawater. Solution: 4. A quantity of 7.85g of a compound having the empirical formula C5H4 is dissolved 301g of benzene. The freezing point of the solution is 1.05 below that of pure benzene. What are the molar mass and molecular formula of this compound?Solution: Since the formula mass of is and the molar mass is found to be , the molecular formula of the compound is .5. Ethylene glycol (EG) CH2(OH)CH2(OH), is a common automobile antifreeze. it is cheap, water-soluble, and fairly nonvolatile (b.p.197).Calculate the freezing point of a solution containing 651g of this substance in 2505g of water. Would you keep this substance in your car radiator during the summer? The molar mass of ethylene glycol is 62.01g.Solution: M(EG)=62.01gmol-1 Because the solution will boil at 102.15,it would be preferable to leave the antifreeze in your car radiator in summer to prevent the solution from boiling.6. A solution is prepared by dissolving 35.0g of hemoglobin (Hb) in enough water to make up one liter in volume.If the osmotic pressure of the solution is found to be 10.0mmHg at 25,calculate the molar mass of hemoglobin.Solution: the concentration of the solution: 7. A 0.86 percent by mass solution of NaCl is called “physiological saline” because its osmotic pressure is equal to that of the