第一篇教材實(shí)驗(yàn)探究類型二物質(zhì)制備實(shí)驗(yàn)4一定溶質(zhì)質(zhì)量分?jǐn)?shù)的氯化鈉溶液的配制探究思路明確任務(wù)(原料和目標(biāo)產(chǎn)品)→分析物質(zhì)性質(zhì)→設(shè)計(jì)路線→選取實(shí)驗(yàn)裝置→確定實(shí)驗(yàn)操作基本方法從溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算出發(fā),進(jìn)行溶液的簡單配制1.配制質(zhì)量分?jǐn)?shù)為6%的氯化鈉溶液藥匙實(shí)驗(yàn)裝置

實(shí)驗(yàn)儀器托盤天平、_______、燒杯、量筒、玻璃棒、____________膠頭滴管3實(shí)驗(yàn)步驟(1)計(jì)算:配制50g質(zhì)量分?jǐn)?shù)為6%的氯化鈉溶液,所需氯化鈉和水的質(zhì)量分別為氯化鈉________g,水______g(體積為______mL,水的密度近似為1g·cm-3)(2)_______:用托盤天平稱量所需的氯化鈉,放入燒杯中(3)量取:用________mL量筒量取所需的水,倒入盛有氯化鈉的燒杯中(4)溶解:用_______攪拌,目的是____________________(5)裝瓶貼標(biāo)簽:將溶液裝入細(xì)口瓶,貼上標(biāo)簽:________________4747稱量50玻璃棒加快氯化鈉溶解NaCl溶液6%2.配制質(zhì)量分?jǐn)?shù)為3%的氯化鈉溶液膠頭滴管實(shí)驗(yàn)儀器燒杯、量筒、___________、___________實(shí)驗(yàn)步驟(1)計(jì)算:配制50g質(zhì)量分?jǐn)?shù)為3%的氯化鈉溶液,所需質(zhì)量分?jǐn)?shù)為6%的氯化鈉溶液和水的質(zhì)量分別為6%的氯化鈉溶液______g(體積24mL),水______mL(2)________

(3)________

(4)裝瓶貼標(biāo)簽玻璃棒2525量取混勻【公式小結(jié)】(1)溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算:溶質(zhì)的質(zhì)量分?jǐn)?shù)＝__________________________________溶劑質(zhì)量＝________________________(2)濃溶液用水稀釋時(shí):①濃溶液中溶質(zhì)的質(zhì)量______(填“＞”“＜”或“＝”)稀溶液中溶質(zhì)的質(zhì)量。②所用水的質(zhì)量＝_________________________?！?00%溶液質(zhì)量-溶質(zhì)質(zhì)量＝稀溶液質(zhì)量-濃溶液質(zhì)量1.用固體試劑配制溶液的誤差分析減少俯視量取溶劑①溶劑的量_______了質(zhì)量分?jǐn)?shù)偏大量取的水在轉(zhuǎn)移到燒杯中時(shí),灑出稱量時(shí),所用砝碼已生銹或沾有油污②溶質(zhì)溶解的量_______了增多左碼右物稱量(當(dāng)用到游碼時(shí))①溶質(zhì)溶解的量變少了質(zhì)量分?jǐn)?shù)偏小溶質(zhì)有撒落固體試劑中含有不溶性雜質(zhì)_____視量取溶劑;燒杯內(nèi)壁上有水②溶劑的量增多了仰2.用濃溶液配制稀溶液的誤差分析實(shí)驗(yàn)結(jié)果質(zhì)量分?jǐn)?shù)偏大質(zhì)量分?jǐn)?shù)偏小誤差分析量取濃溶液時(shí)仰視讀數(shù);量取水時(shí)俯視讀數(shù)量取濃溶液時(shí)俯視讀數(shù);量取水時(shí)仰視讀數(shù);燒杯不干燥;稀釋過程中,有濃溶液灑出用質(zhì)量分?jǐn)?shù)為8%的硝酸鉀溶液配制50g質(zhì)量分?jǐn)?shù)為4%的硝酸鉀溶液。(1)需要8%的硝酸鉀溶液的質(zhì)量為______g。(2)該溶液配制過程中除用到量筒、玻璃棒和膠頭滴管外,還需要的儀器是__________。25燒杯(3)已知KNO3在20℃時(shí)的溶解度是31.6g,此溫度下可以將15.8gKNO3固體溶于水配成恰好飽和的溶液________g,該溶液通電一段時(shí)間后得到溶液的質(zhì)量為47.8g,過濾后得到KNO3固體______(結(jié)果精確到0.1)g。65.84.3(4)用硝酸鉀固體配制一定溶質(zhì)質(zhì)量分?jǐn)?shù)的溶液時(shí),下列操作會導(dǎo)致溶液中溶質(zhì)質(zhì)量分?jǐn)?shù)偏小的是______