60℃時(shí),將24gKNO3溶于96g水中,所得溶液的質(zhì)量分?jǐn)?shù)為_(kāi)_____,現(xiàn)此溶液等分為四份:
(1)第一份溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為_(kāi)_____;
(2)取第二份溶液,將其溫度升到100℃,此時(shí)溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為_(kāi)_____;
(3)取第三份,使其溶質(zhì)的質(zhì)量分?jǐn)?shù)變?yōu)樵瓉?lái)的2倍,需加KNO3固體______g或蒸發(fā)______g
(4)取第四份,使其溶質(zhì)的質(zhì)量分?jǐn)?shù)變?yōu)樵瓉?lái)的一半,需加入水______g.
解:60℃時(shí),將24 g硝酸鉀溶于96 g水中,所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為:
×100%=20%
將溶液分成四等份時(shí),溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)不變.每份溶液中溶質(zhì)的質(zhì)量為6g,溶液的質(zhì)量為30g.
(1)溶液具有均一穩(wěn)定性所以第一份溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)是20%.
(2)將溶液升高溫度時(shí),因溶質(zhì)和溶劑都沒(méi)有變化,所以溶質(zhì)質(zhì)量分?jǐn)?shù)不變?nèi)匀皇?0%.
(3)使其溶質(zhì)質(zhì)量分?jǐn)?shù)變?yōu)樵瓉?lái)的2倍,可通過(guò)加入溶質(zhì)或蒸發(fā)溶劑進(jìn)行,此時(shí)溶質(zhì)質(zhì)量分?jǐn)?shù)變?yōu)?0%.
則設(shè)加入硝酸鉀的質(zhì)量為x
×100%=40%
x=10g
設(shè)蒸發(fā)水的質(zhì)量為y
×100%=40%
y=15g
(4)使其溶質(zhì)質(zhì)量分?jǐn)?shù)變?yōu)樵瓉?lái)的一半,即溶質(zhì)的質(zhì)量分?jǐn)?shù)為10%
設(shè)加入的水的質(zhì)量為Z
×100%=10%
z=30g.
故答案為:20%;(1)20%(2)20%(3)10;15(4)30
分析:溶液是均一、穩(wěn)定的混合物,所以將溶液分成三等份時(shí),溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)不變.然后再根據(jù)溶質(zhì)的質(zhì)量分?jǐn)?shù)計(jì)算公式進(jìn)行計(jì)算就行了.
點(diǎn)評(píng):將溶液均分時(shí),因溶液具有均一、穩(wěn)定的特點(diǎn),所以每份溶液中溶質(zhì)和溶劑都是原來(lái)的四分之一.所以在計(jì)算時(shí)不能代錯(cuò)了數(shù)值.