224ML某气体在标准状况下的质量为0.32g，气体的摩尔质量是多少
标况下，224ml气体（既0.224l）的物质的量n=0.224/22.4=0.01mol根据n=m/m
得m=m/n=0.32g/0.01mol=32g/mol
由此知该气体是氧气o2
绛旓細A 鎽╁皵璐ㄩ噺鐨勫崟浣嶄负g路mol-1锛岀浉瀵瑰垎瀛愯川閲忕殑鍗曚綅涓1.鏍囧噯鐘跺喌涓锛224mL姘斾綋涓0.01mol锛岃姘斾綋鐨鎽╁皵璐ㄩ噺涓 =32g路mol-1鏁呴堿銆傛湰棰樿娉ㄦ剰鍖哄垎鎽╁皵璐ㄩ噺涓庣浉瀵瑰垎瀛愯川閲忕殑鍗曚綅銆
绛旓細鏍囧喌涓锛224ml姘斾綋锛堟棦0.224l锛夌殑鐗╄川鐨勯噺n=0.224/22.4=0.01mol 鏍规嵁n=m/m 寰梞=m/n=0.32g/0.01mol=32g/mol 鐢辨鐭ヨ姘斾綋鏄哀姘攐2
绛旓細鍦ㄦ爣鍑嗙姸鍐典笅224mL鏌愭皵浣鐨勭墿璐ㄧ殑閲忎负0.01mol 璇ユ皵浣撶殑鎽╁皵璐ㄩ噺M=0.64/0.01=64g/mol 璇ユ皵浣撶殑鐩稿鍒嗗瓙璐ㄩ噺涓64
绛旓細鏍囧噯鐘舵佷笅涓鎽╁皵浠讳綍姘斾綋鐨勪綋绉兘鏄22.4L 锛屾墍浠224mL姘斾綋鐨勬懇灏旀暟涓 0.224/22.4=0.01mol 銆傚姝ゅ彲浠ユ眰鍑烘懇灏旇川閲=0.64/0.01=64锛屾墍浠ヨ姘斾綋鐩稿鍒嗗瓙璐ㄩ噺涓64
绛旓細锛1锛塶=VVm=0.224L/22.4L?mol-1=0.01 mol锛孧=mn=0.40g/0.01mol=40 g/mol锛岀瓟锛氭懇灏旇川閲忎负40 g/mol锛涳紙2锛夎闇VL鐨0.5mol/L AlCl3鏍规嵁 n=C脳V锛宯锛圕l-锛=1 mol/L脳2脳0.5L=0.5mol/L脳3脳VV=23L锛屾晠绛旀涓猴細23锛涳紙3锛塶锛圕uSO4锛=CVn锛圕uSO4锛=2.0mol/L脳0.1L=0...
绛旓細鏍囧喌涓嬩綋绉224ml鍗0.224L鍗0.01mol,0.64g/0.01mol,鎵浠ユ爣鍐典笅姝姘斾綋鎽╁皵閲忎负64g/mol,
绛旓細0.1g鐨勭櫧鑹叉矇娣,涓虹⒊閰搁挋杩欐槸鍙互鑲畾鐨勫叾鍚噺涓0.1g闄や簬100g/mol鍗0.001mol.鐢变簬姘㈡哀鍖栭挋鏄0.1L*0.02mol/L=0.002mol 鎵浠ユ湁涓ょ鎯呭喌 1.姘㈡哀鍖栭挋杩囬噺锛0.001mol纰抽吀閽欏嵆涓轰簩姘у寲纰崇殑閲忥紝涔熷氨鏄敳鐑风殑閲忎负0.001mol.鎵浠ョ敳鐑风殑浣撶Н涓22.4ml,鍗犳讳綋绉殑1/10,鎵浠ョ敳鐑峰拰姘ф皵鐨勪綋绉瘮涓1/...
绛旓細鏍囧噯鐘跺喌涓,22.4L浠讳綍姘斾綋鐨鐗╄川鐨勯噺鍧囦负1mol.鎵浠,22.4mL=0.001mol 璇ユ皵浣撶殑鎽╁皵璐ㄩ噺涓猴細0.32g/0.001mol=320g/mol 鏍规嵁浣犳彁渚涚殑鏁版嵁鍙兘杩欐牱璁＄畻銆傝妫鏌ユ暟鎹槸鍚︽纭
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