ÌâÄ¿ÄÚÈÝ

19£®£¨1£©ÂËÒºAÔÚ100¡æ¾­Õô·¢Å¨ËõÖÁÌå»ýΪ30.0mL£¬ÀäÈ´ÖÁ10¡æºó£¬Îö³öNa2S2O3•5H2O¾§Ì壮ÈôÒª¼ÆËãËùµÃµ½µÄNa2S2O3•5H2OµÄÖÊÁ¿£¬ÄãÈÏΪABD£¨ÌîдѡÏî×Öĸ£©
A¡¢»¹ÐèÒªÌṩ10¡æÊ±Na2S2O3µÄÈܽâ¶ÈΪ£¨60.0g£©
B¡¢»¹ÐèÒªÌṩ100¡æÊ±ÈÜÒºµÄÃܶȣ¨1.12g•cm-3£©
C¡¢»¹ÐèÒªÌṩ½á¾§ºóÊ£ÓàÈÜÒºµÄÌå»ý£¨10.0mL£©
D¡¢»¹ÐèÒªÌṩNa2SO3ÓëS·ÛÍêÈ«·´Ó¦ºóÉú³ÉµÄNa2S2O3µÄÖÊÁ¿£¨19.1g£©
£¨2£©¸ù¾ÝµÚ£¨1£©Ð¡ÌâÄãµÄÑ¡Ôñ£¨Ñ¡ÓÃÆäÊý¾Ý£©£¬¼ÆËãÀäÈ´ÖÁ10¡æºó£¬Îö³öNa2S2O3•5H2O¾§ÌåµÄÖÊÁ¿Îª24.8g£®

·ÖÎö ¸ù¾ÝÒÑÓеÄ֪ʶ½øÐзÖÎö½â´ð£¬Òª¼ÆËãËùµÃµ½Na2S2O3•5H2OµÄÖÊÁ¿£¬ÐèÒªÖªµÀ10¡æÊ±Na2S2O3µÄÈܽâ¶È£¬ÒÔ¼°100¡æÊ±ÈÜÒºµÄÃܶȺÍNa2SO3ÓëS·ÛÍêÈ«·´Ó¦ºóÉú³ÉµÄNa2S2O3µÄÖÊÁ¿£¬¾Ý´Ë½â´ð¼´¿É£®

½â´ð ½â£º£¨1£©ÐèÒªÖªµÀ10¡æÊ±Na2S2O3µÄÈܽâ¶È£¬ÒÔ¼°100¡æÊ±ÈÜÒºµÄÃܶȺÍNa2SO3ÓëS·ÛÍêÈ«·´Ó¦ºóÉú³ÉµÄNa2S2O3µÄÖÊÁ¿£¬¹ÊÌABD£»
£¨2£©ÈÜÒºÖÐË®µÄÖÊÁ¿=30£®0mL¡Á1£®12g/mL-19.1g=14.5g£¬ÉèÎö³öµÄNa2S2O3•5H2OµÄÖÊÁ¿Îªx£¬Ôò£º$\frac{60.0g}{100g}=\frac{19.1g-\frac{158g/mol}{248g/mol}¡Áx}{14.5g-\frac{90g/mol}{248g/mol}¡Áx}$
½âµÃx=24.8g£¬
¹ÊÌ24.8g£®

µãÆÀ ±¾Ì⿼²éµÄÊǽᾧµÄ֪ʶ£¬Íê³É´ËÌ⣬ÐèÒªÒÀ¾ÝÈܽâ¶ÈµÄ֪ʶ½øÐУ¬ÄѶȽϴó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø