Question

現(xiàn)有t1℃時(shí)某溶液l00g，溶質(zhì)質(zhì)量分?jǐn)?shù)為20%，溶液密度為1.5g/mL，降溫至t2℃時(shí)有5g晶體（不含結(jié)晶水）析出，則t2℃時(shí)溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)為（　�。�

A．25.4%

B．20.3%

C．15.8%

D．5.9%

Answer 1

分析：根據(jù)溶液的質(zhì)量和溶質(zhì)的質(zhì)量分?jǐn)?shù)來計(jì)算溶質(zhì)和溶劑的質(zhì)量，再根據(jù)降溫時(shí)溶質(zhì)的質(zhì)量減少，而溶劑質(zhì)量不變來計(jì)算t2℃時(shí)溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)．

解答：解：由t1℃時(shí)某溶液l00g，溶質(zhì)質(zhì)量分?jǐn)?shù)為20%，
則溶質(zhì)的質(zhì)量為100g×20%=20g，
溶劑的質(zhì)量為100g-20g=80g，
再由降溫至t2℃時(shí)有5g晶體析出，且晶體中不含結(jié)晶水，
則t2℃時(shí)溶液中的溶質(zhì)質(zhì)量為20g-5g=15g，
則t2℃時(shí)溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)為

15g

80g+15g

×100%=15.8%，
故選C．

點(diǎn)評(píng)：本題考查溶質(zhì)質(zhì)量分?jǐn)?shù)的簡單計(jì)算，明確降溫時(shí)溶劑的質(zhì)量不變，變化的只有溶質(zhì)的質(zhì)量是解答的關(guān)鍵，然后利用溶質(zhì)的質(zhì)量分?jǐn)?shù)的計(jì)算方法來解答即可．