Question

（2005?荔灣區(qū)二模）在100g稀硫酸中加入20%的硫酸溶液50g，使溶液的質(zhì)量分?jǐn)?shù)為12%，則原稀硫酸的溶質(zhì)的質(zhì)量分?jǐn)?shù)為（　　）

A．2%

B．4%

C．8%

D．10%

Answer 1

分析：在100g稀硫酸中加入20%的硫酸溶液50g，使溶液的質(zhì)量分?jǐn)?shù)為12%，則混合后所得溶液質(zhì)量為兩溶液的質(zhì)量和、溶液中溶質(zhì)質(zhì)量為兩溶液所含溶質(zhì)質(zhì)量和，利用溶液中溶質(zhì)質(zhì)量分?jǐn)?shù)計(jì)算公式，計(jì)算所加硫酸的溶質(zhì)質(zhì)量分?jǐn)?shù)．

解答：解：設(shè)所加硫酸的溶質(zhì)質(zhì)量分?jǐn)?shù)為x
100g?x+50g×20%=（50g+100g）×12%
解得：x=8%
分析所給選項(xiàng)可以知道選項(xiàng)C是正確的．
故選C．

點(diǎn)評(píng)：根據(jù)溶液組成特點(diǎn)，利用溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)計(jì)算公式，解決不同質(zhì)量分?jǐn)?shù)的溶液混合問(wèn)題，正確判斷溶液的組成是解決問(wèn)題的基礎(chǔ)．