ÎÞ»ú¼°·ÖÎö»¯Ñ§()Ï°Ìâ½â´ð

ÎÞ»ú¼°·ÖÎö»¯Ñ§(¢ñ)Ï°Ìâ½â´ð

Ŀ ¼

µÚ1Õ ·ÖÉ¢Ìåϵ¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­1 µÚ2Õ µÚ3Õ µÚ4Õ µÚ5Õ µÚ6Õ µÚ7Õ µÚ8Õ µÚ9Õ µÚ10ÕµÚ11ÕµÚ12ÕµÚ13ÕµÚ14ÕµÚ15ÕµÚ16ÕµÚ17ÕµÚ18Õ»¯Ñ§ÈÈÁ¦Ñ§»ù´¡¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­8 »¯Ñ§·´Ó¦ËÙÂʺͻ¯Ñ§Æ½ºâ¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­15 ÎïÖʽṹ¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­22 ·ÖÎö»¯Ñ§¸ÅÊö¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­27 Ëá¼îƽºâ¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­31 Ëá¼îµÎ¶¨·¨¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­36 ³ÁµíÈܽâƽºâ¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­44 ÖØÁ¿·ÖÎö·¨ºÍ³ÁµíµÎ¶¨·¨¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­49 Åäλ»¯ºÏÎï¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­53 ÅäλµÎ¶¨·¨¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­62 Ñõ»¯»¹Ô­·´Ó¦¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­67 Ñõ»¯»¹Ô­µÎ¶¨·¨¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­78 µçÊÆ·ÖÎö·¨¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­83 Îü¹â¹â¶È·ÖÎö·¨¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­86 ÊÔÑù·ÖÎöÖг£ÓõķÖÀë·½·¨¼ò½é¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­90 ÖØÒªÉúÃüÔªËؼòÊö¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­93 Ô­×Ӻ˻¯Ñ§¼ò½é¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­96

µÚ1Õ ·ÖÉ¢Ìåϵ

1-1£®ÈËÌå×¢ÉäÓõÄÉúÀíÑÎË®ÖУ¬º¬ÓÐNaCl 0.900%£¬ÃܶÈΪ1.01g?mL-1£¬ÈôÅäÖÆ´ËÈÜÒº3.00¡Á103g£¬ÐèNaCl¶àÉÙ¿Ë£¿¸ÃÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈÊǶàÉÙ£¿

½â£ºÅäÖƸÃÈÜÒºÐèNaClµÄÖÊÁ¿Îª£º m(NaCl)=0.900%¡Á3.00¡Á103g=27.0g ¸ÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ£º

27.0g/58.5g?mol-1-1 c(NaCl)==0.155mol?L3-1-3(3.00?10g/1.01g?mL)?101-2£®°Ñ30.0gÒÒ´¼(C2H5OH)ÈÜÓÚ50.0gËÄÂÈ»¯Ì¼(CCl4)ÖÐËùµÃÈÜÒºµÄÃܶÈΪ1.28g?mL-1£¬¼ÆË㣺(1)ÒÒ´¼µÄÖÊÁ¿·ÖÊý£»(2)ÒÒ´¼µÄÎïÖʵÄÁ¿Å¨¶È£»(3)ÒÒ´¼µÄÖÊÁ¿Ä¦¶ûŨ¶È£»(4)ÒÒ´¼µÄĦ¶û·ÖÊý¡£

½â£º

(1)w(C2H5OH) =

m(C2H5OH)30.0g=0.38 ?m(C2H5OH)?m(CCl4)30.0g?50.0gn(C2H5OH)30.0g/46g?mol-1(2)c(C2H5OH)==10.4mol¡¤L-1 ?-1-3V[(30.0+50.0)g/1.28g?mL]?1030.0g-1n(CHOH)-146g?mol25(3)b(C2H5OH) ==13.0mol¡¤kg ?-3m(CCl4)50.0?10kg30.0gn(C2H5OH)46g?mol-1(4)x(C2H5OH) = =30.0g50.0gn(C2H5OH)+n(CCl4)+-146g?mol153.6g?mol-1=

0.65?0.66

0.65?0.331-3£®½«5.0g NaOH¡¢NaCl¡¢CaCl2·Ö±ðÖÃÓÚË®ÖУ¬Åä³É500mLÈÜÒº£¬ÊÔÇóc(NaOH)¡¢c(NaCl)¡¢c(

1CaCl2)¡£ 2½â£ºNaOH¡¢NaClºÍ

1CaCl2µÄĦ¶ûÖÊÁ¿·Ö±ðΪ£º 2 1

M(NaOH)= 40g¡¤mol-1 M(NaCl)= 58.5g¡¤mol-1 M(ËùÒÔ£¬ËüÃǵÄÎïÖʵÄÁ¿Å¨¶È·Ö±ðΪ£º c(NaOH) =c(NaCl) =c(

1CaCl2)= 55.5g¡¤mol-1 25.0/40= 0.25 (mol¡¤L-1)

500/10005.0/58.5= 0.17 (mo¡¤L-1)

500/100015.0/55.5CaCl2)== 0.18 (mol¡¤L-1) 2500/10001-4£®ÑÎËẬHCl 37.0%(ÖÊÁ¿·ÖÊý)£¬ÃܶÈΪ1.19g¡¤mL-1¡£¼ÆË㣺 (1)ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶È¡£ (2)ÑÎËáµÄÖÊÁ¿Ä¦¶ûŨ¶È¡£ (3)HClºÍH2OµÄÎïÖʵÄÁ¿·ÖÊý¡£ ½â£º(1)ÑÎËáµÄ·Ö×ÓÁ¿Îª36.5 c(HCl) =

1000?1.19?37.0%= 12.06 mol¡¤L-1

36.5(2)b(HCl) =

1000?37.0%= 16.09 mol¡¤kg-1

36.5?(1?37.0%)(3)x2 =

37.0%/36.5=0.225£»

37.0%/36.5?(1?37.0%)/18x1 = 1£­x2 = 0.775

1-5£®¼ÆËã0.10mol¡¤L-1K3[Fe(CN)6]ÈÜÒºµÄÀë×ÓÇ¿¶È¡£ ½â£ºI=

1(0.30¡Á12+0.10¡Á32)=0.6 21-6£®Ó¦Óõ°ݣ­Ðݿ˶û¼«ÏÞ¹«Ê½¼ÆËã0.10mol¡¤L-1KClÈÜÒºÖеÄÀë×Óƽ¾ù»î¶ÈϵÊý¡£ ½â£ºKCl I=0.1

lg????0.509z??z?1?II=?0.509?1?0.1=-0.12 1?0.1¦Ã¡À=0.76

1-7£®½«19gijÉúÎï¼îÈÜÓÚ100gË®ÖУ¬²âµÃÈÜÒºµÄ·ÐµãÉý¸ßÁË0.060K£¬Äý¹Ìµã½µµÍÁË

2

0.220K¡£¼ÆËã¸ÃÉúÎï¼îµÄÏà¶Ô·Ö×ÓÖÊÁ¿¡£

½â£º?Tb?KbbB?0.512K?kg?mol?1?M = 1.62¡Á103 g¡¤mol-1

19g?0.060K

M?0.1kg?Tf?KfbB?1.86K?kg?mol?1?M=1.61¡Á103 g¡¤mol-1

19g=0.220K

M?0.1kgËùÒÔ£¬¸ÃÉúÎï¼îµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª£º

(1.61?1.62)?103g?mol-1?1.62?103g?mol-1

21-8£®Èܽâ0.324gÁòÓÚ4.00g C6H6ÖУ¬Ê¹C6H6µÄ·ÐµãÉÏÉýÁË0.81K¡£ÎÊ´ËÈÜÒºÖеÄÁò·Ö×ÓÊÇÓɼ¸¸öÔ­×Ó×é³ÉµÄ£¿[Kb(C6H6)=2.53 K?kg?mol-1]

½â£ºÉè´ËÈÜÒºÖеÄÁò·Ö×ÓÓÉx¸öÁòÔ­×Ó×é³É£¬ÔòÆäĦ¶ûÖÊÁ¿Îª32x g?mol-1

0.324gnB32xg?mol?12.53?1 bB???mol?kg?3mA4?10xÓÉÓÚ?Tb?K·ÐbB£¬

?Tb?0.81K?2.53K?kg?mol-1?ËùÒÔx=8

2.53mol?kg-1 x¹ÊÈÜÒºÖеÄ1¸öÁò·Ö×ÓÊÇÓÉ8¸öÁòÔ­×Ó×é³ÉµÄ¡£

1-9£®¼ÆËã0.005mol¡¤L-1KClÈÜÒºÔÚ398KʱµÄÉø͸ѹ£º(1)ÓÃŨ¶È¼ÆË㣻(2)Óûî¶È¼ÆËã(¦Ã¡À=0.92)¡£

½â£º(1)¦°=icRT=2¡Á0.005¡Á8.314¡Á398=33.09(kPa)

(2)¦°=i¦Ã¡ÀcRT=2¡Á0.92¡Á0.005¡Á8.314¡Á398=30.44(kPa)

1-10£®101mgÒȵºËØÈÜÓÚ10.0mLË®£¬¸ÃÈÜÒºÔÚ25.0¡æʱµÄÉø͸ѹÊÇ4.34kPa£¬¼ÆËãÒȵºËصÄĦ¶ûÖÊÁ¿ºÍ¸ÃÈÜÒºµÄÄý¹Ìµã¡£

½â£ºÉèÒȵºËصÄĦ¶ûÖÊÁ¿ÎªM£¬ÓÉÓÚÉø͸ѹ¿ÉÓÉÏÂʽ¼ÆËã ¦°=cBRT

3