課題3溶質(zhì)的質(zhì)量分?jǐn)?shù)第1課時(shí)溶質(zhì)的質(zhì)量分?jǐn)?shù)教學(xué)分析:通過(guò)前面的學(xué)習(xí),學(xué)生已經(jīng)知道溶液是由溶質(zhì)和溶劑組成的,本課題是從定量角度讓學(xué)生進(jìn)一步認(rèn)識(shí)溶液的組成,為今后學(xué)習(xí)離子反應(yīng)打下基礎(chǔ)。另外,本節(jié)課中涉及溶液的稀釋、溶液中溶質(zhì)質(zhì)量分?jǐn)?shù)的基本計(jì)算,這些知識(shí)可以提高學(xué)生的計(jì)算能力和分析解決實(shí)際問(wèn)題的能力。教學(xué)目標(biāo):1.通過(guò)配制不同濃度的硫酸銅溶液,認(rèn)識(shí)溶質(zhì)的質(zhì)量分?jǐn)?shù)的概念及意義。2.通過(guò)理解溶質(zhì)質(zhì)量分?jǐn)?shù)定義,能進(jìn)行有關(guān)溶質(zhì)的質(zhì)量分?jǐn)?shù)和有關(guān)溶液的稀釋、增濃問(wèn)題的簡(jiǎn)單計(jì)算。教學(xué)重點(diǎn):溶質(zhì)質(zhì)量分?jǐn)?shù)的含義和簡(jiǎn)單計(jì)算。教學(xué)難點(diǎn):有關(guān)溶液的稀釋、增濃問(wèn)題的計(jì)算。教學(xué)用具:硫酸銅、氯化鈉、水、燒杯、玻璃棒。教學(xué)思路:教學(xué)過(guò)程教師活動(dòng)學(xué)生活動(dòng)設(shè)計(jì)意圖【導(dǎo)入】播放死海的視頻【提問(wèn)】死海中的鹽濃度怎樣定量表示呢?觀(guān)看視頻,思考并回答相關(guān)問(wèn)題。通過(guò)死海不死這一異常的情境,導(dǎo)入新課,激發(fā)學(xué)生的學(xué)習(xí)興趣。任務(wù)一溶質(zhì)的質(zhì)量分?jǐn)?shù)活動(dòng)1探究溶質(zhì)質(zhì)量分?jǐn)?shù)的含義1.探究有色溶液顏色的深淺與溶液濃度的關(guān)系進(jìn)行不同濃度的硫酸銅溶液的配制實(shí)驗(yàn)。燒杯編號(hào)溶液顏色比較溶劑質(zhì)量/g溶質(zhì)質(zhì)量/g溶液質(zhì)量/gⅠⅡⅢ【提問(wèn)】(1)哪個(gè)燒杯中溶液最濃、哪個(gè)燒杯中溶液最稀,你是如何判斷的?(2)在施用農(nóng)藥時(shí),準(zhǔn)確知道藥液的溶質(zhì)質(zhì)量分?jǐn)?shù)有什么意義?2.溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算公式溶質(zhì)的質(zhì)量分?jǐn)?shù)=溶質(zhì)質(zhì)量溶液質(zhì)量3.溶質(zhì)質(zhì)量分?jǐn)?shù)的含義每100g某溶液中溶質(zhì)的質(zhì)量。如某氯化鈉溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為16%,其含義是100g該溶液中氯化鈉的質(zhì)量為16g。4.計(jì)算溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)時(shí)應(yīng)注意什么?【歸納】(1)溶質(zhì)的質(zhì)量指的是形成溶液的那部分溶質(zhì)的質(zhì)量,未溶解的不計(jì)算在內(nèi)。(2)計(jì)算時(shí)質(zhì)量單位要統(tǒng)一,不要忘記×100%,最后結(jié)果要寫(xiě)成百分?jǐn)?shù)。(3)溶液具有穩(wěn)定性,只要外界條件不變,其溶質(zhì)的質(zhì)量分?jǐn)?shù)不變。進(jìn)行實(shí)驗(yàn),填表。觀(guān)察思考,回答問(wèn)題:(1)燒杯Ⅲ中溶液最濃,燒杯Ⅰ中溶液最稀;從溶液顏色的深淺可以判斷。(2)意義:既可以有效地殺蟲(chóng)滅菌又不會(huì)毒害農(nóng)作物。交流、討論,回答問(wèn)題。通過(guò)直觀(guān)實(shí)驗(yàn)及現(xiàn)象、精確的數(shù)字,使學(xué)生理解溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)的表示方法,并理解溶質(zhì)質(zhì)量分?jǐn)?shù)與濃度的關(guān)系。任務(wù)二配制不同質(zhì)量分?jǐn)?shù)的氯化鈉溶液活動(dòng)2配制不同質(zhì)量分?jǐn)?shù)的氯化鈉溶液1.展示下圖配制過(guò)程。按圖配制氯化鈉溶液,填表。通過(guò)兩種不同濃度溶液的配制,使學(xué)生學(xué)會(huì)計(jì)算溶質(zhì)的質(zhì)量分?jǐn)?shù);通過(guò)氯化鈉的溶解度與飽和溶液中的溶質(zhì)質(zhì)量分?jǐn)?shù),使學(xué)生學(xué)會(huì)飽和溶液中溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算。填寫(xiě)表格。溶質(zhì)質(zhì)量/g溶劑(水)質(zhì)量/g固體能否全部溶解溶質(zhì)的質(zhì)量分?jǐn)?shù)109020802.飽和溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)已知20℃時(shí),氯化鈉的溶解度是36g。有人說(shuō)“20℃時(shí)氯化鈉飽和溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為36%”,這種說(shuō)法對(duì)嗎?為什么?歸納飽和溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算公式:飽和溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)=溶解度溶解度+思考,回答。任務(wù)三溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算活動(dòng)3直接利用公式計(jì)算1.把10g氯化鈉放入40g水中,完全溶解后,溶液中氯化鈉的質(zhì)量分?jǐn)?shù)是多少?解:溶質(zhì)的質(zhì)量分?jǐn)?shù)=溶質(zhì)質(zhì)量溶液質(zhì)量×100%所得溶液的質(zhì)量=溶質(zhì)的質(zhì)量+溶劑的質(zhì)量=10g+40g=50g。所得溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)=10g50g×100%=20%答:溶液中氯化鈉的質(zhì)量分?jǐn)?shù)是20%。2.在20℃時(shí),將40g氯化鈉放入100g水中,結(jié)果有4g固體剩余,求溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)。解:溶質(zhì)的質(zhì)量分?jǐn)?shù)=溶質(zhì)質(zhì)量溶液質(zhì)量×100%溶解的氯化鈉的質(zhì)量=40g-4g=36g。所得溶液的質(zhì)量=36g+100g=136g。所得溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)=36g136g×100%=26.5%答:溶液中氯化鈉的質(zhì)量分?jǐn)?shù)是26.5%?！局v解】未溶解的溶質(zhì)不能計(jì)算在溶液中溶質(zhì)的質(zhì)量中,計(jì)算溶液的質(zhì)量時(shí)也應(yīng)減去未溶解的溶質(zhì)的質(zhì)量。【練習(xí)】例1:在農(nóng)業(yè)生產(chǎn)中,常需要用質(zhì)量分?jǐn)?shù)為16%的氯化鈉溶液來(lái)選種?，F(xiàn)要配制150kg這種溶液,需要氯化鈉和水的質(zhì)量各是多少?傾聽(tīng)。答題:解:溶質(zhì)的質(zhì)量分?jǐn)?shù)=溶質(zhì)質(zhì)量溶液質(zhì)量×100%溶質(zhì)質(zhì)量=溶液質(zhì)量×溶質(zhì)的質(zhì)量分?jǐn)?shù)=150kg×16%=24kg通過(guò)例題的計(jì)算,理解溶液中各個(gè)量的含義及它們之間的關(guān)系,能用溶質(zhì)質(zhì)量分?jǐn)?shù)解決簡(jiǎn)單的化學(xué)問(wèn)題?！局v解】利用公式時(shí)注意公式的變形:溶質(zhì)的質(zhì)量分?jǐn)?shù)=溶質(zhì)質(zhì)量溶液質(zhì)量溶質(zhì)質(zhì)量=溶液質(zhì)量×溶質(zhì)的質(zhì)量分?jǐn)?shù);溶液質(zhì)量=溶質(zhì)質(zhì)量溶質(zhì)的質(zhì)量分?jǐn)?shù)溶液質(zhì)量=溶質(zhì)質(zhì)量+溶劑質(zhì)量;溶液質(zhì)量=溶液體積×溶液密度?；顒?dòng)4探究溶液的稀釋的計(jì)算例2:化學(xué)實(shí)驗(yàn)室現(xiàn)在有98%的濃硫酸,但在實(shí)驗(yàn)中常需要用較稀的硫酸溶液。要把50g質(zhì)量分?jǐn)?shù)為98%的濃硫酸稀釋為質(zhì)量分?jǐn)?shù)為20%的硫酸溶液,需要水的質(zhì)量是多少?【講解】溶液稀釋前后,溶質(zhì)的質(zhì)量不變?；顒?dòng)5探究溶液的增濃的計(jì)算例3:(1)有40g溶質(zhì)的質(zhì)量分?jǐn)?shù)為5%的氯化鈣溶液,若將其溶質(zhì)的質(zhì)量分?jǐn)?shù)增大到10%,求需加入氯化鈣的質(zhì)量?！局v解】增加溶質(zhì)后,溶質(zhì)的質(zhì)量、溶液的質(zhì)量都會(huì)增加。溶劑質(zhì)量=溶液質(zhì)量-溶質(zhì)質(zhì)量=150kg-24kg=126kg答:需要24kg氯化鈉和126kg水。答題:解:設(shè)稀釋后溶液的質(zhì)量為x。50g×98%=x×20%x=50g×98%=245g需要水的質(zhì)量=245g-50g=195g答:需要水的質(zhì)量是195g。交流、討論,歸納溶液稀釋和增濃問(wèn)題的計(jì)算方法,理解各個(gè)量的含義及它們之間的關(guān)系。答題:(1)解:設(shè)需加入氯化鈣的質(zhì)量為x。40g×5%+x40g+x≈2.22g答:需加入氯化鈣的質(zhì)量為2.22g。講練結(jié)合,使學(xué)生深刻理解溶液中各個(gè)量在稀釋或增濃時(shí)會(huì)發(fā)生怎樣的變化,體會(huì)溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算在實(shí)際生產(chǎn)、生活中的應(yīng)用。(2)有40g溶質(zhì)質(zhì)量分?jǐn)?shù)為5%的氯化鈣溶液,若將其溶質(zhì)的質(zhì)量分?jǐn)?shù)增大到10%,求需蒸發(fā)掉水的質(zhì)量?！局v解】蒸發(fā)了水,溶液的質(zhì)量會(huì)減少。(3)有40g溶質(zhì)質(zhì)量分?jǐn)?shù)為5%的氯化鈣溶液,若將其溶質(zhì)的質(zhì)量分?jǐn)?shù)增大到10%,需加入15%的氯化鈣溶液的質(zhì)量為多少?【講解】加入濃溶液后,溶質(zhì)和溶液的質(zhì)量都會(huì)增加。(2)解:設(shè)需蒸發(fā)掉水的質(zhì)量為y。40g×5%40g-y=20g答:需蒸發(fā)掉水的質(zhì)量為20g。(3)解:設(shè)需加入15%的氯化鈣溶液的質(zhì)量為z。40g×5%+zz=40g答:需加入15%的氯化鈣溶液的質(zhì)量為40g。課堂小結(jié)【總結(jié)板書(shū)】學(xué)生討論并歸納總結(jié)。學(xué)會(huì)歸納總結(jié)并構(gòu)建模型。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